2.2 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words, ∆ x v = lim (2.1a) ∆ t → 0 ∆ t Fig. 2.1 Determining velocity from position-time d x = (2.1b) graph. Velocity at t = 4 s is the slope of the dt tangent to the graph at that instant. lim where the symbol stands for the operation ∆→t 0 Now, we decrease the value of ∆t from 2 s to 1 of taking limit as ∆tg0 of the quantity on its s. Then line P1P2 becomes Q1Q2 and its slope right. In the language of calculus, the quantity gives the value of the average velocity over on the right hand side of Eq. (2.1a) is the the interval 3.5 s to 4.5 s. In the limit ∆t → 0, differential coefficient of x with respect to t and the line P1P2 becomes tangent to the position- d x time curve at the point P and the velocity at t is denoted by (see Appendix 2.1). It is the d t = 4 s is given by the slope of the tangent at rate of change of position with respect to time, that point. It is difficult to show this process graphically. But if we useat that instant. numerical method to obtain the value of We can use Eq. (2.1a) for obtaining the the velocity, the meaning of the limiting value of velocity at an instant either process becomes clear. For the graph shown graphically or numerically. Suppose that we in Fig. 2.1, x = 0.08 t3. Table 2.1 gives the want to obtain graphically the value of value of ∆x/∆t calculated for ∆t equal to 2.0 s, velocity at time t = 4 s (point P) for the motion 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = of the car represented in Fig.2.1 calculation. 4.0 s. The second and third columns give theLet us take ∆t = 2 s centred at t = 4 s. Then, t  t    ∆ ∆by the definition of the average velocity, the t + t −     and t 2 = and the value of t1= 2 2slope of ( Fig. 2.1) gives the value of     line P1P2 average velocity over the interval 3 s to 5 s. fourth and the fifth columns give the ∆x Table 2.1 Limiting value of at t = 4 s ∆ t Reprint 2025-26 MOTION IN A STRAIGHT LINE 15 3 a + 16b – a – 4b corresponding values of x, i.e. x (t1) = 0.08 t1 = = 6.0 × b 2.0 -1 ⊳and x (t2) = 0.08 t23. The sixth column lists the = 6.0 × 2.5 =15 m s difference ∆x = x (t2) – x (t1) and the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting + 24.0 m s–1 and a velocity of – 24.0 m s–1 — value 3.84 m s–1 which is the value of velocity at both have an associated speed of 24.0 m s-1. It should be noted that though average speed over dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt magnitude of the average velocity, manner, we can calculate velocity at each instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a 2.3 ACCELERATIONconvenient method. For this, we must carefully plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the ⊳ increasing distance of fall. This led to the Example 2.1 The position of an object concept of acceleration as the rate of change moving along x-axis is given by x = a + bt2 of velocity with time. where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time interval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? v 2 – v1 ∆v (2.2)Answer In notation of differential calculus, the a = = t 2 – t1 ∆tvelocity is where v2 and v1 are the instantaneous velocities dx d 2 -1 2b t = 5.0 t m s or simply velocities at time t2 and t1 . It is thev = = ( a + bt ) = dt dt average change of velocity per unit time. The SI At t = 0 s, v = 0 m s–1 and at t = 2.0 s, unit of acceleration is m s–2 . v = 10 m s-1 . On a plot of velocity versus time, the average acceleration is the slope of the straight line x ( 4.0 ) − x ( 2.0 )Average velocity = connecting the points corresponding to (v2, t2) 4.0 − 2.0 and (v1, t1). Reprint 2025-26 16 PHYSICS Instantaneous acceleration is defined in the (c) An object is moving in negative direction same way as the instantaneous velocity : with a negative acceleration. ∆v d v (d) An object is moving in positive direction lim a = = (2.3) till time t1, and then turns back with the d t ∆→ t 0 ∆ t same negative acceleration. The acceleration at an instant is the slope of the tangent to the v–t curve at that An interesting feature of a velocity-time instant. graph for any moving object is that the area Since velocity is a quantity having both under the curve represents the magnitude and direction, a change in displacement over a given time interval. A velocity may involve either or both of these general proof of this statement requires use of factors. Acceleration, therefore, may result calculus. We can, however, see that it is true from a change in speed (magnitude), a for the simple case of an object moving with change in direction or changes in both. Like constant velocity u. Its velocity-time graph is velocity, acceleration can also be positive, as shown in Fig. 2.4. negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vo at t = 0 and v at time t, we have v − v 0 a = or, v = v 0 + a t (2.4) t − 0 Fig. 2.3 Velocity–time graph for motions with Fig. 2.2 Position-time graph for motion with constant acceleration. (a) Motion in positive (a) positive acceleration; (b) negative direction with positive acceleration, acceleration, and (c) zero acceleration. (b) Motion in positive direction with Let us see how velocity-time graph looks like negative acceleration, (c) Motion in for some simple cases. Fig. 2.3 shows velocity- negative direction with negative acceleration, (d) Motion of an object withtime graph for motion with constant acceleration negative acceleration that changesfor the following cases : direction at time t1. Between times 0 to (a) An object is moving in a positive direction t1, it moves in positive x - direction with a positive acceleration. and between t1 and t2 it moves in the (b) An object is moving in positive direction opposite direction. with a negative acceleration. Reprint 2025-26 MOTION IN A STRAIGHT LINE 17 Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which Fig. 2.5 Area under v-t curve for an object with is the displacement in this time interval. How uniform acceleration. come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at As explained in the previous section, the area the answer. under v-t curve represents the displacement. Therefore, the displacement x of the object is : Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp 1 x = ( v – v 0 ) t + v 0 t (2.5)kinks at some points implying that the 2 functions are not differentiable at these But v − v 0 = a tpoints. In any realistic situation, the functions will be differentiable at all points 1 2 Therefore, x = a t + v 0 tand the graphs will be smooth. 2 What this means physically is that 1 2 or, x = v 0 t + at (2.6)acceleration and velocity cannot change 2 values abruptly at an instant. Changes are Equation (2.5) can also be written as always continuous. v + v 0 x = t = v t (2.7a)2.4 KINEMATIC EQUATIONS FOR 2 UNIFORMLY ACCELERATED MOTION where, For uniformly accelerated motion, we can derive some simple equations that relate displacement v + v 0 v = (constant acceleration only)(x), time taken (t), initial velocity (v0), final 2 velocity (v) and acceleration (a). Equation (2.4) (2.7b) already obtained gives a relation between final and initial velocities v and v0 of an object moving Equations (2.7a) and (2.7b) mean that the object with uniform acceleration a : has undergone displacement x with an average velocity equal to the arithmetic average of the v = v0 + at (2.4) initial and final velocities. From Eq. (2.4), t = (v – v0)/a. Substituting this in This relation is graphically represented in Fig. 2.5. Eq. (2.7a), we get The area under this curve is : Area between instants 0 and t = Area of triangle  v + v 0   v − v 0  v 2 − v 02 x = v t =ABC + Area of rectangle OACD  2   a = 2a 1 2 2 = (v – v 0 ) t + v 0 t v = v 0 + 2ax (2.8) 2 Reprint 2025-26 18 PHYSICS This equation can also be obtained by t v 0 + at ) d tsubstituting the value of t from Eq. (2.4) into Eq. = ∫ 0 ( (2.6). Thus, we have obtained three important equations : 1 2 x – x 0 = v 0 t + a t 2 v = v 0 + at 1 2 1 2 x = x 0 + v 0 t + a t x = v 0t + at 2 2 We can write v 2 = v 02 + 2ax (2.9a) d v d v d x d v a = = = v d t d x d t d x connecting five quantities v0, v, a, t and x. These or, v dv = a dxare kinematic equations of rectilinear motion for Integrating both sides,constant acceleration. The set of Eq. (2.9a) were obtained by v x v d v = a d xassuming that at t = 0, the position of the particle, ∫ v 0 ∫ x 0 x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non- v 2 – v 02 = a ( x – x 0 ) zero, say x0. Then Eqs. (2.9a) are modified 2 (replacing x by x – x0 ) to : 2 2 v = v 0 + 2a ( x – x 0 ) v = v 0 + at The advantage of this method is that it can be used 1 2 for motion with non-uniform acceleration x = x 0 + v 0t + at (2.9b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (2.9c) important cases. ⊳ ⊳ ⊳ Example 2.3 A ball is thrown vertically Example 2.2 Obtain equations of motion upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how longAnswer By definition will it be before the ball hits the ground? d v Take g = 10 m s–2. a = d t dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the v t ∫ v 0 d v = ∫ 0 a d t ground, as shown in Fig. 2.6. Now vo = + 20 m s–1, t d t (a is a = – g = –10 m s–2, = a ∫ 0 v = 0 m s–1 constant) If the ball rises to height y from the point of v – v 0 = at launch, then using the equation 2 + 2 a 0 ( y – y 0 ) v = v 0 + at v 2 = v we get d x Further, v = 0 = (20)2 + 2(–10)(y – y0) d t Solving, we get, (y – y0) = 20 m. dx = v dt Integrating both sides (b) We can solve this part of the problem in two x t ways. Note carefully the methods used. ∫ x 0 dx = ∫v0 d t Reprint 2025-26 MOTION IN A STRAIGHT LINE 19 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. ⊳ ⊳ Example 2.4 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 2.6 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use v = 0 – g t = –9.8 t m s–1equation y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y 0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t2 2 the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, with time have been plotted in Fig. 2.7(a), (b)Therefore, the total time taken by the ball before and (c). it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y = 0 m vo = 20 m s-1, a = –10m s–2, t = ? (a) Reprint 2025-26 20 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 2 y = −1 gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 2.2. If we take (–1/ 2) gτ2 as y0 — the position coordinate after first time interval τ, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. ⊳ ⊳ Example 2.6 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and theFig. 2.7 Motion of an object under free fall. braking capacity, or deceleration, –a that (a) Variation of acceleration with time. (b) Variation of velocity with time. is caused by the braking. Derive an (c) Variation of distance with time ⊳ expression for stopping distance of a vehicle in terms of vo and a. ⊳ Example 2.5 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling 2 motion v2 = vo + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. – v 02 d s = 2aAnswer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals τ and find out the distances the square of the initial velocity. Doubling the Table 2.2 Reprint 2025-26 MOTION IN A STRAIGHT LINE 21 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example, in school zones. ⊳ ⊳ Example 2.7 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think Fig. 2.8 Measuring the reaction time. and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before Answer The ruler drops under free fall. he slams the brakes of the car is the Therefore, vo = 0, and a = – g = –9.8 m s–2. The reaction time. Reaction time depends distance travelled d and the reaction time tr are on complexity of the situation and on related by an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between Or, your thumb and forefinger (Fig. 2.8). Given d = 21.0 cm and g = 9.8 m s–2 the reaction After you catch it, find the distance d time is travelled by the ruler. In a particular case, d was found to be 21.0 cm. ⊳ Estimate reaction time. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small : ∆ x d x v = lim v = lim = ∆→t 0 ∆→t 0 ∆t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. Reprint 2025-26 22 PHYSICS 3. Average acceleration is the change in velocity divided by the time interval during which the change occurs : ∆ v a = ∆t 4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆v d v a = lim a = lim = ∆→t 0 ∆→t 0 ∆ t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 5. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at 1 2 x = v0 t + at 2 v 2 = v 02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x – x0). Reprint 2025-26 MOTION IN A STRAIGHT LINE 23 POINTS TO PONDER 1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. Reprint 2025-26 24 PHYSICS EXERCISES 2.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice). Fig. 2.9 2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? 2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). 2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. Reprint 2025-26 MOTION IN A STRAIGHT LINE 25 2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. 2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. 2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] Fig. 2.10 2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. 2.13 Figure 2.11shows the x-t plot of one- dimensional motion of a particle. Is it correct to say from the graph that the particle moves Fig. 2.11 in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car). Reprint 2025-26 26 PHYSICS 2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12): Fig. 2.12 2.16 Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s. Fig. 2.13 2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 2.14 2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Fig. 2.15 Reprint 2025-26 CHAPTER THREE MOTION IN A PLANE 3.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 3.1 Introduction describe the motion of an object along a straight line. We 3.2 Scalars and vectors found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two3.3 Multiplication of vectors by real numbers directions are possible. But in order to describe motion of an 3.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 3.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to3.6 Vector addition — analytical method add, subtract and multiply vectors ? What is the result of 3.7 Motion in a plane multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and3.8 Motion in a plane with constant acceleration acceleration in a plane. We then discuss motion of an object 3.9 Projectile motion in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail3.10 Uniform circular motion the projectile motion. Circular motion is a familiar class of Summary motion that has a special significance in daily-life situations. Points to ponder We shall discuss uniform circular motion in some detail. Exercises The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 3.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided Reprint 2025-26 28 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of

9.3 Alkenes Alkenes are unsaturated hydrocarbons Fig. 9.4 Orbital picture of ethene depictingcontaining at least one double bond. What σ bonds only should be the general formula of alkenes? If there is one double bond between two carbon 9.3.2 Nomenclature atoms in alkenes, they must possess two For nomenclature of alkenes in IUPAC system, hydrogen atoms less than alkanes. Hence, the longest chain of carbon atoms containing general formula for alkenes is CnH2n. Alkenes the double bond is selected. Numbering of the are also known as olefins (oil forming) since chain is done from the end which is nearer to Reprint 2025-26 Hydrocarbons 307 Fig. 9.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ Solutionof alkanes. It may be remembered that first member of alkene series is: CH2 (replacing (i) 2,8-Dimethyl-3, 6-decadiene; n by 1 in CnH2n) known as methene but has (ii) 1,3,5,7 Octatetraene; a very short life. As already mentioned, first (iii) 2-n-Propylpent-1-ene; stable member of alkene series is C2H4 known (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes Problem 9.8 are given below : Calculate number of sigma (σ) and pi (π) Structure IUPAC name bonds in the above structures (i-iv). CH3 – CH = CH2 Propene SolutionCH3 – CH2 – CH = CH2 But – l - ene σ bonds : 33, π bonds : 2CH3 – CH = CH–CH3 But-2-ene σ bonds : 17, π bonds : 4CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 23, π bond : 1CH2 = C – CH3 2-Methylprop-1-ene | σ bonds : 41, π bond : 1 CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene 9.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 9.7 (C2H4) and propene (C3H6) can have only one Write IUPAC names of the following structure but alkenes higher than propene compounds: have different structures. Alkenes possessing (i) (CH3)2CH – CH = CH – CH2 – CH C4H8 as molecular formula can be written in  the following three ways: CH3 – CH – CH | I. 1 2 3 4 C2H5 CH2 = CH – CH2 – CH3 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 3 4 | | CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 | But-2-ene CH3 (C4H8) Reprint 2025-26 308 chemistry III. 1 2 3 In (a), the two identical atoms i.e., both CH2 = C – CH3 the X or both the Y lie on the same side | of the double bond but in (b) the two X or CH3 two Y lie across the double bond or on the 2-Methylprop-1-ene opposite sides of the double bond. This (C4H8) results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in Structures I and III, and II and III are the two arrangements is different. Therefore,the examples of chain isomerism whereas they are stereoisomers. They would have thestructures I and II are position isomers. same geometry if atoms or groups around C=C bond can be rotated but rotation around Problem 9.9 C=C bond is not free. It is restricted. For Write structures and IUPAC names of understanding this concept, take two pieces different structural isomers of alkenes of strong cardboards and join them with the corresponding to C5H10. help of two nails. Hold one cardboard in your Solution one hand and try to rotate the other. Can (a) CH2 = CH – CH2 – CH2 – CH3 you really rotate the other cardboard ? The answer is no. The rotation is restricted. This Pent-1-ene illustrates that the restricted rotation of atoms (b) CH3 – CH=CH – CH2 – CH3 or groups around the doubly bonded carbon Pent-2-ene atoms gives rise to different geometries of such compounds. The stereoisomers of this (c) CH3 – C = CH – CH3 type are called geometrical isomers. The | isomer of the type (a), in which two identical CH3 atoms or groups lie on the same side of the 2-Methylbut-2-ene double bond is called cis isomer and the (d) CH3 – CH – CH = CH2 other isomer of the type (b), in which identical | atoms or groups lie on the opposite sides of CH3 the double bond is called trans isomer . Thus 3-Methylbut-1-ene cis and trans isomers have the same structure but have different configuration (arrangement (e) CH2 = C – CH2 – CH3 of atoms or groups in space). Due to different | arrangement of atoms or groups in space, CH3 these isomers differ in their properties like 2-Methylbut-1-ene melting point, boiling point, dipole moment, solubility etc. Geometrical or cis-trans isomersGeometrical isomerism: Doubly bonded of but-2-ene are represented below :carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that Reprint 2025-26 Hydrocarbons 309 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two (iii) C6H5CH = CH – CH3forms as given below from which it is clear that in the trans-but-2-ene, the two methyl (iv) CH3CH = CCl CH3 groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus Solution making the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. 9.3.4 Preparation 1. From alkynes: Alkynes on partial reduction with calculated amount of cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised (µ = 0.33D) (µ = 0) charcoal partially deactivated with poisons like sulphur compounds or quinoline give In the case of solids, it is observed that the alkenes. Partially deactivated palladisedtrans isomer has higher melting point than charcoal is known as Lindlar’s catalyst.the cis form. Alkenes thus obtained are having cis Geometrical or cis-trans isomerism geometry. However, alkynes on reductionis also shown by alkenes of the types with sodium in liquid ammonia form transXYC = CXZ and XYC = CZW alkenes. Problem 9.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (9.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (9.31) iii) CH≡ CH+H2 Pd/C CH2 =CH2 (9.32) Ethyne Ethene CH3–C≡ CH+H2 Pd/C CH3–CH =CH2 iv) Propyne Propene (9.33) Will propene thus obtained show Problem 9.11 geometrical isomerism? Think for the reason in support of your answer. Which of the following compounds will show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on (i) (CH3)2C = CH – C2H5 heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, Reprint 2025-26 310 chemistry ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is β-carbon atom. known as dehydrohalogenation i.e., removal of halogen acid. This is example of β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (9.37) 9.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. (9.34) The first three members are gases, the next fourteen are liquids and the higher ones are Nature of halogen atom and the alkyl group solids. Ethene is a colourless gas with a faint determine rate of the reaction. It is observed sweet smell. All other alkenes are colourless that for halogens, the rate is: iodine > and odourless, insoluble in water but fairly bromine > chlorine, while for alkyl groups soluble in non-polar solvents like benzene, it is : tertiary > secondary > primary. petroleum ether. They show a regular increase 3. From vicinal dihalides: Dihalides in in boiling point with increase in size i.e., every which two halogen atoms are attached – CH2 group added increases boiling point by to two adjacent carbon atoms are known 20–30 K. Like alkanes, straight chain alkenes as vicinal dihalides. Vicinal dihalides on have higher boiling point than isomeric treatment with zinc metal lose a molecule branched chain compounds. of ZnX2 to form an alkene. This reaction Chemical properties is known as dehalogenation. Alkenes are the rich source of loosely held CH2Br–CH2Br + Zn CH2=CH2+ ZnBr2 pi (π) electrons, due to which they show (9.35) addition reactions in which the electrophiles add on to the carbon-carbon double bond toCH3CHBr–CH2Br + Zn CH3CH=CH2 form the addition products. Some reagents +ZnBr2 also add by free radical mechanism. There (9.36) are cases when under special conditions, alkenes also undergo free radical substitution4. From alcohols by acidic dehydration: reactions. Oxidation and ozonolysis reactions You have read during nomenclature of are also quite prominent in alkenes. A brief different homologous series in Unit 12 description of different reactions of alkenes that alcohols are the hydroxy derivatives is given below: of alkanes. They are represented by R–OH where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add with concentrated sulphuric acid form up one molecule of dihydrogen gas in alkenes with the elimination of one water the presence of finely divided nickel, molecule. Since a water molecule is palladium or platinum to form alkanes eliminated from the alcohol molecule in (Section 9.2.2) the presence of an acid, this reaction is 2. Addition of halogens : Halogens like known as acidic dehydration of alcohols. bromine or chlorine add up to alkene to This reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under Reprint 2025-26 Hydrocarbons 311 normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation (9.42) which you will study in higher classes. Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding (9.38) molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov (9.39) rule can be better understood in terms of mechanism of the reaction. 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes Mechanism to form alkyl halides. The order of Hydrogen bromide provides an electrophile, reactivity of the hydrogen halides is H +, which attacks the double bond to form HI > HBr > HCl. Like addition of halogens carbocation as shown below : to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes (a) less stable (b) more stableAddition reactions of HBr to symmetrical primary carbocation secondary carbocationalkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates CH2=CH2+H–Br CH3–CH2–Br (9.40) because it is formed at a faster rate. (ii) The carbocation (b) is attacked by Br– ionCH3–CH=CH–CH3+HBr CH3–CH–CHCH3 to form the product as follows : Br (9.41) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two 2-Bromopropane possible products are I and II. (major product) Reprint 2025-26 312 chemistry Anti Markovnikov addition or peroxide effect or Kharash effect In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl The secondary free radical obtained in the and Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 the primary. This explains the formation of at the University of Chicago. This reaction 1-bromopropane as the major product. It may is known as peroxide or Kharash effect be noted that the peroxide effect is not observed or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due rule. to the fact that the H–Cl bond being (C6H5CO)2O2 stronger (430.5 kJ mol –1) than H–Br bond CH3 – CH=CH2+HBr CH3–CH2 (363.7 kJ mol –1), is not cleaved by the free radical, whereas the H–I bond is weaker CH2Br (296.8 kJ mol –1) and iodine free radicals 1–Bromopropane combine to form iodine molecules instead of adding to the double bond. (9.43) Mechanism : Peroxide effect proceeds via Problem 9.12 free radical chain mechanism as given below: Write IUPAC names of the products obtained by addition reactions of HBr to(i) hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution Homolysis C. 6H5+H–Br C6H3+ B. r(ii) 4. Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. Reprint 2025-26 Hydrocarbons 313 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (9.49) KMnO4/H+ CH3 – CH=CH–CH3 2CH3COOH (9.44) But-2-ene Ethanoic acid (9.50) 7. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H2O to smaller molecules. This reaction is highly useful in detecting the position of the double (9.45) bond in alkenes or other unsaturated compounds.5. Addition of water : In the presence of a few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (9.51) (9.46) 6. Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. (9.52) 8. Polymerisation: You are familiar with (9.47) polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called (9.48) polymers. This reaction is known as b) Acidic potassium permanganate or acidic polymerisation. The simple compounds potassium dichromate oxidises alkenes to from which polymers are made are called Reprint 2025-26 314 chemistry monomers. Other alkenes also undergo are named as derivatives of the corresponding polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. n(CH2 =CH2) High temp./pressureCatalyst —( CH2–CH2 )— The position of the triple bond is indicated by the first triply bonded carbon. Common Polythene and IUPAC names of a few members of alkyne (9.53) series are given in Table 9.2. High temp./pressure You have already learnt that ethyne and n(CH3 –CH=CH2) Catalyst —( CH–CH2 )—n propyne have got only one structure but there are two possible structures for butyne – CH3 (i) but-1-yne and (ii) but-2-yne. Since these Polypropene two compounds differ in their structures (9.54) due to the position of the triple bond, they Polymers are used for the manufacture of plastic are known as position isomers. In how bags, squeeze bottles, refrigerator dishes, toys, many ways, you can construct the structure pipes, radio and T.V. cabinets etc. Polypropene for the next homologue i.e., the next alkyne is used for the manufacture of milk crates, with molecular formula C5H8? Let us try to plastic buckets and other moulded articles. arrange five carbon atoms with a continuous Though these materials have now become chain and with a side chain. Following are the common, excessive use of polythene and possible structures : polypropylene is a matter of great concern for Structure IUPAC name all of us. 1 2 3 4 5 I. HC≡ C– CH2– CH2– CH3 Pent–1-yne

5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳