Relative Motion — Relative velocity in 1D and 2D
Kinematics
13
JEE Qs
8%
Hard
75
min
Always draw a clear vector diagram and resolve all velocities into perpendicular components before applying relative motion equations in 2D problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1Relative velocity vec(v)_AB signifies the velocity of object A as observed from the reference frame of object B (or with respect to B). The fundamental relation is a vector subtraction: vec(v)_AB = vec(v)_A - vec(v)_B.
- 2For problems involving an agent moving within a medium (e.g., boat in river, airplane in wind), the velocity of the agent relative to the medium is usually provided. The key is to correctly use vector addition: vec(v)_agent_ground = vec(v)_agent_medium + vec(v)_medium_ground.
- 3In 2D problems, always draw clear vector diagrams and resolve velocities into perpendicular components (e.g., x and y axes) before applying relative velocity equations.
- 4Minimum distance problems between two moving objects can often be simplified by considering the motion of one object relative to the other (i.e., making one object stationary in the relative frame) and then finding the perpendicular distance from the 'stationary' object to the 'moving' object's relative path.
- 5Carefully distinguish between 'shortest time' and 'shortest path' scenarios in river-boat problems, as they require different vector orientations for the boat's velocity relative to the river.
⚠️ Common Mistakes
- ✕Incorrectly applying vector subtraction by simply subtracting magnitudes or using wrong directions without resolving components, especially in 2D.
- ✕Confusing the different frames of reference, particularly in river-boat or rain-man problems (e.g., using velocity of boat relative to river as its velocity relative to ground directly).
- ✕Failing to draw clear vector diagrams, leading to sign errors or misinterpretation of angles and directions.
- ✕Not understanding that relative velocity is a vector, and its direction is crucial; it's not always along the line connecting the two objects.
📝 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
- Class 11 Physics Ch 4: Motion in a Plane