Equations of Motion — v=u+at, s=ut+½at², v²=u²+2as
Kinematics
13
JEE Qs
8%
Hard
60
min
Always draw a simple diagram and establish a clear, consistent sign convention for all vector quantities (displacement, velocity, acceleration) before attempting to solve any problem.
🧮 Key Formulas
✅ Key Points for JEE
- 1These equations are strictly applicable ONLY for motion with CONSTANT acceleration.
- 2Establish a consistent sign convention (e.g., upward positive, downward negative) for displacement, velocity, and acceleration at the start of every problem.
- 3Understand the physical meaning of each variable: u (initial velocity), v (final velocity), a (constant acceleration), s (displacement), t (time interval).
- 4Distance travelled in the nth second (s_n) is a specific application for finding displacement during that particular one-second interval.
- 5Choose the most appropriate equation based on the given and required variables to simplify calculations.
⚠️ Common Mistakes
- ✕Applying these equations when acceleration is variable; they are only valid for constant acceleration.
- ✕Incorrectly using sign conventions for vector quantities (displacement, velocity, acceleration), especially when direction changes.
- ✕Confusing 's' (displacement) with 'distance travelled' when the object reverses direction.
- ✕Failing to recognize that 'at rest' or 'comes to a stop' implies zero velocity (u=0 or v=0).
📝 Practice Questions
See allQ27.A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ? (1) (KE) (2) (KE) 8 2 (3) (KE) (4) (KE) 16 4
Q28.The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s ? (1) 30 m (2) 11 m (3) 10 m (4) 13 m
Q50.A particle is projected at an angle of 30∘ from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1 . The ratio h0 : h1 is _________ [Take, g = 10 m/s2 ]
Q43.The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km . (1) 12 (2) 3 (3) 6 (4) 9
Q27.The position vector of a moving body at any instant of time is given as →r = . The magnitude (5t2^i −5t^j)m and direction of velocity at t = 2 s is, (1) 5√15 m/s, making an angle of tan−1 4 with - ve (2) 5√15 m/s, making an angle of tan−1 4 with + ve Y axis X axis (3) 5√17 m/s, making an angle of tan−1 4 with + ve (4) 5√17 m/s, making an angle of tan−1 4 with - ve X axis Y axis
Q31.Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+ α), respectively, with the horizontal direction. The ratio of their maximum heights attained is : (1) 1−tan α (2) 1−sin 2α 1+tan α 1+sin 2α (3) 1+sin 2α (4) 1+sin α 1−sin 2α 1−sin α
NCERT Chapters
- Class 11 Physics Ch 3: Motion in a Straight Line