Concept Card: Relative Velocity in One Dimension (1D Kinematics)
1. Definition of Relative Velocity:
The velocity of a body $A$ as observed from the reference frame of another body $B$ is defined as the relative velocity of $A$ with respect to $B$ ($\vec{v}_{AB}$):
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B$
Similarly, the velocity of $B$ relative to $A$ is:
$\vec{v}_{BA} = \vec{v}_B – \vec{v}_A = -\vec{v}_{AB}$
where $\vec{v}_A$ and $\vec{v}_B$ are the velocities measured relative to a common reference frame (the ground).
2. One-Dimensional Sign Conventions:
Choosing one direction as positive ($+x$):
- Moving in the Same Direction: If both bodies move along $+x$ with speeds $v_A$ and $v_B$:
$v_{AB} = v_A – v_B$
The relative speed is the difference of their individual speeds. - Moving in Opposite Directions: If $A$ moves along $+x$ and $B$ moves along $-x$:
$v_{AB} = v_A – (-v_B) = v_A + v_B$
The relative speed is the sum of their individual speeds (approaching or receding).
3. Relative Acceleration in 1D:
$\vec{a}_{AB} = \vec{a}_A – \vec{a}_B$
Fundamental Theorem for Free Fall:
If two particles $A$ and $B$ move vertically under gravity alone:
$\vec{a}_A = -g\hat{j}, \quad \vec{a}_B = -g\hat{j} \implies \vec{a}_{AB} = (-g) – (-g) = 0$
The relative acceleration between two freely falling bodies is always strictly zero. Hence, the relative motion of one freely falling body with respect to another is always uniform motion with constant relative velocity!
4. Relative Kinematic Equations (When $a_{\text{rel}} = \text{constant}$):
- $v_{\text{rel}} = u_{\text{rel}} + a_{\text{rel}} t$
- $s_{\text{rel}} = u_{\text{rel}} t + \frac{1}{2}a_{\text{rel}} t^2$
- $v_{\text{rel}}^2 = u_{\text{rel}}^2 + 2 a_{\text{rel}} s_{\text{rel}}$
5. High-Yield JEE Problem Types:
- Trains Crossing and Overtaking:
For two trains of lengths $L_1$ and $L_2$ to completely cross each other, the relative distance to be traversed is $s_{\text{rel}} = L_1 + L_2$.- Opposite directions: $t_{\text{cross}} = \frac{L_1 + L_2}{v_1 + v_2}$
- Same direction ($v_1 > v_2$): $t_{\text{overtake}} = \frac{L_1 + L_2}{v_1 – v_2}$
- Rear-End Collision Avoidance:
If Vehicle 1 (initial speed $u_1$, deceleration $a_1$) trails Vehicle 2 (speed $u_2$, deceleration $a_2$) with initial gap $d$:
Relative initial velocity: $u_{\text{rel}} = u_1 – u_2$.
Relative deceleration: $a_{\text{rel}} = a_1 – a_2$.
Collision is avoided if the relative stopping distance does not exceed $d$:
$d_{\text{min}} = \frac{(u_1 – u_2)^2}{2(a_1 – a_2)} \le d$ - Escalator Kinematics:
Let $L$ be the escalator length. If a person walks up a stationary escalator in time $t_1$ ($v_p = L/t_1$) and the moving escalator alone carries a standing person in time $t_2$ ($v_e = L/t_2$), then walking up the moving escalator takes time $T$:
$v_{\text{net}} = v_p + v_e \implies \frac{L}{T} = \frac{L}{t_1} + \frac{L}{t_2} \implies T = \frac{t_1 t_2}{t_1 + t_2}$.
Solved Examples
Example 1 (Trains Crossing in Same and Opposite Directions):
Two trains $A$ and $B$ on parallel tracks have lengths $L_A = 150\text{ m}$ and $L_B = 100\text{ m}$. Train $A$ travels at $v_A = 72\text{ km/h}$ ($20\text{ m/s}$) and Train $B$ travels at $v_B = 36\text{ km/h}$ ($10\text{ m/s}$).
Find the time taken for the trains to completely pass each other when:
(a) They move in opposite directions.
(b) They move in the same direction.
Solution:
The total relative distance to be cleared for complete crossing is the sum of their lengths:
$s_{\text{rel}} = L_A + L_B = 150\text{ m} + 100\text{ m} = 250\text{ m}$.
(a) Opposite directions:
Relative speed: $v_{\text{rel}} = v_A + v_B = 20\text{ m/s} + 10\text{ m/s} = 30\text{ m/s}$.
$t_{\text{cross}} = \frac{s_{\text{rel}}}{v_{\text{rel}}} = \frac{250\text{ m}}{30\text{ m/s}} = \frac{25}{3}\text{ s} \approx 8.33\text{ seconds}$.
(b) Same direction:
Relative speed: $v_{\text{rel}} = v_A – v_B = 20\text{ m/s} – 10\text{ m/s} = 10\text{ m/s}$.
$t_{\text{overtake}} = \frac{s_{\text{rel}}}{v_{\text{rel}}} = \frac{250\text{ m}}{10\text{ m/s}} = 25\text{ seconds}$.
Example 2 (Collision Avoidance Between Two Vehicles):
A car traveling at $u_1 = 90\text{ km/h}$ ($25\text{ m/s}$) is trailing a truck moving at $u_2 = 54\text{ km/h}$ ($15\text{ m/s}$) in the same direction on a straight highway. When the distance between them is $d = 50\text{ m}$, the car driver applies brakes producing a constant deceleration $a_1 = 1.5\text{ m/s}^2$, while the truck accelerates forward at $a_2 = 0.5\text{ m/s}^2$.
(a) Determine whether a collision occurs.
(b) Find the closest distance of approach between the two vehicles.
Solution:
Work in the frame of reference of the truck:
Initial relative velocity: $u_{\text{rel}} = u_1 – u_2 = 25 – 15 = +10\text{ m/s}$.
Relative acceleration: $a_{\text{rel}} = a_1 – a_2 = (-1.5) – (+0.5) = -2.0\text{ m/s}^2$.
Initial separation: $d = 50\text{ m}$.
The closest approach occurs when the relative velocity drops to zero ($v_{\text{rel}} = 0$):
$v_{\text{rel}}^2 = u_{\text{rel}}^2 + 2 a_{\text{rel}} s_{\text{rel}} \implies 0 = 10^2 + 2(-2.0) s_{\text{rel}}$
$4 s_{\text{rel}} = 100 \implies s_{\text{rel}} = 25\text{ m}$.
(a) Since the maximum relative forward displacement $s_{\text{rel}} = 25\text{ m} < 50\text{ m}$, the car comes to relative rest before reaching the truck. Hence, no collision occurs.
(b) Closest distance of approach: $d_{\text{closest}} = d – s_{\text{rel}} = 50\text{ m} – 25\text{ m} = 25\text{ meters}$.
Example 3 (Two Vertical Projectiles with Launch Time Delay):
Ball $A$ is projected vertically upward from the ground with speed $u_A = 40\text{ m/s}$ at $t = 0$. Exactly $\Delta t = 2.0\text{ s}$ later, Ball $B$ is projected vertically upward along the same vertical line with speed $u_B = 60\text{ m/s}$. Taking $g = 10\text{ m/s}^2$:
(a) Find the time after Ball $A$’s launch when the two balls collide.
(b) Find the height above the ground at which the collision occurs.
Solution:
(a) Let $t$ be the time measured from the launch of Ball $A$.
Then Ball $B$ travels for time $(t – 2)$ for $t \ge 2\text{ s}$.
Heights of both balls:
$y_A(t) = u_A t – \frac{1}{2}gt^2 = 40t – 5t^2$
$y_B(t) = u_B(t – 2) – \frac{1}{2}g(t – 2)^2 = 60(t – 2) – 5(t – 2)^2$
Collision occurs when $y_A = y_B$:
$40t – 5t^2 = 60t – 120 – 5(t^2 – 4t + 4) = 60t – 120 – 5t^2 + 20t – 20 = 80t – 140 – 5t^2$.
Canceling $-5t^2$ from both sides:
$40t = 80t – 140 \implies 40t = 140 \implies t = \frac{140}{40} = 3.5\text{ seconds}$.
(b) Height of collision:
$y(3.5) = 40(3.5) – 5(3.5^2) = 140 – 61.25 = 78.75\text{ meters}$.
Example 4 (Escalator Kinematics):
A commuter walking up a stationary escalator takes $t_1 = 60\text{ seconds}$ to reach the top. When standing still on the moving escalator, the commuter is carried to the top in $t_2 = 90\text{ seconds}$.
(a) Determine the time taken if the commuter walks up the moving escalator.
(b) Determine the time taken if the commuter walks down the upward moving escalator at the same walking speed.
Solution:
Let $L$ be the length of the escalator.
Walking speed: $v_w = \frac{L}{t_1} = \frac{L}{60}$.
Escalator speed: $v_e = \frac{L}{t_2} = \frac{L}{90}$.
(a) Walking up the moving escalator:
$v_{\text{up}} = v_w + v_e = \frac{L}{60} + \frac{L}{90} = L\left(\frac{3 + 2}{180}\right) = \frac{5L}{180} = \frac{L}{36}$.
$T_{\text{up}} = \frac{L}{v_{\text{up}}} = \frac{L}{L/36} = 36\text{ seconds}$.
(b) Walking down the upward moving escalator:
$v_{\text{down}} = v_w – v_e = \frac{L}{60} – \frac{L}{90} = L\left(\frac{3 – 2}{180}\right) = \frac{L}{180}$.
$T_{\text{down}} = \frac{L}{v_{\text{down}}} = \frac{L}{L/180} = 180\text{ seconds}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
Two cars $A$ and $B$ are traveling towards each other on a straight road with speeds $20\text{ m/s}$ and $30\text{ m/s}$ respectively. If their initial separation is $500\text{ m}$, they will meet after a time of:
(A) $5\text{ s}$
(B) $10\text{ s}$
(C) $20\text{ s}$
(D) $25\text{ s}$
Problem 2 (JEE Main – Single Correct):
A train $150\text{ m}$ long is moving North at $10\text{ m/s}$. A bird flies South at $5\text{ m/s}$ parallel to the train track. The time taken by the bird to cross the train is:
(A) $10\text{ s}$
(B) $15\text{ s}$
(C) $30\text{ s}$
(D) $12\text{ s}$
Problem 3 (JEE Main – Single Correct):
A bus starts from rest with constant acceleration $a = 1.0\text{ m/s}^2$. At the same instant, a man $48\text{ m}$ behind the bus starts running with constant speed $v$ towards the bus. The minimum value of $v$ so that the man can catch the bus is:
(A) $4.8\text{ m/s}$
(B) $9.8\text{ m/s}$
(C) $10.0\text{ m/s}$
(D) $4\sqrt{6}\text{ m/s} \approx 9.8\text{ m/s}$
Problem 4 (JEE Main – Single Correct):
A police car moving on a highway with a speed of $30\text{ km/h}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of $192\text{ km/h}$. If the muzzle speed of the bullet is $150\text{ m/s}$, with what speed does the bullet hit the thief’s car?
(A) $105\text{ m/s}$
(B) $120\text{ m/s}$
(C) $140\text{ m/s}$
(D) $95\text{ m/s}$
Problem 5 (JEE Main – Single Correct):
Two bodies are dropped from different heights $h_1$ and $h_2$ at the same instant. As they fall freely under gravity, the separation between them:
(A) Increases continuously with time
(B) Decreases continuously with time
(C) Remains constant with time
(D) First increases and then decreases
Problem 6 (JEE Advanced – One or More Correct):
Two particles $A$ and $B$ are projected vertically upwards from the ground at the same instant with speeds $u_A = 20\text{ m/s}$ and $u_B = 30\text{ m/s}$. Neglecting air resistance:
(A) The relative velocity $v_B – v_A$ is independent of time and equals $10\text{ m/s}$ upwards until $A$ lands.
(B) The relative acceleration between them is zero.
(C) The separation between the particles increases linearly with time as long as both are in flight.
(D) When $A$ is at its highest point, the distance of $B$ above $A$ is $20\text{ m}$.
Problem 7 (JEE Advanced – One or More Correct):
A passenger on a train moving with uniform velocity $v$ throws a ball vertically upwards with speed $u$ inside a closed carriage. Which of the following statements is/are correct?
(A) To the passenger inside the carriage, the path of the ball is a straight vertical line.
(B) To an observer on the ground, the path of the ball is a parabola.
(C) The time of flight of the ball is identical in both the train and ground reference frames.
(D) The maximum height reached relative to the floor of the train is $\frac{u^2}{2g}$.
Problem 8 (JEE Advanced – One or More Correct):
Two cars $A$ and $B$ move along the same straight line in the same direction. Car $A$ is moving with speed $u_A$ and decelerating with $a_A$. Car $B$ is ahead by distance $d$, moving with speed $u_B$ ($u_A > u_B$) and decelerating with $a_B$. Which of the following conditions guarantee that no collision occurs?
(A) If $a_A \ge a_B$ and $\frac{(u_A – u_B)^2}{2(a_A – a_B)} \le d$
(B) If $a_A = a_B$ and $u_A = u_B$
(C) If the relative velocity drops to zero before the separation vanishes
(D) If $a_A < a_B$, a collision can always be avoided regardless of $d$
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Two trains $A$ and $B$, each of length $100\text{ m}$, are moving on parallel tracks towards each other with uniform speeds $54\text{ km/h}$ and $36\text{ km/h}$. The time taken by them to completely cross each other is $T\text{ seconds}$. Find $T$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A man walks up a stationary escalator in $90\text{ seconds}$. When standing on the same moving escalator, he is carried up in $60\text{ seconds}$. If he walks up the moving escalator, he will reach the top in $t\text{ seconds}$. Find the value of $t$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (D) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 8 | 10. 36
Solution 1:
Relative speed of approach: $v_{\text{rel}} = v_A + v_B = 20 + 30 = 50\text{ m/s}$.
Initial separation: $d = 500\text{ m}$.
Time of meeting: $t = \frac{d}{v_{\text{rel}}} = \frac{500\text{ m}}{50\text{ m/s}} = 10\text{ seconds}$.
Correct Answer: (B)
Solution 2:
Relative speed of bird with respect to train: $v_{\text{rel}} = v_{\text{train}} + v_{\text{bird}} = 10 + 5 = 15\text{ m/s}$.
Distance to be covered: $L = 150\text{ m}$.
Time taken: $t = \frac{L}{v_{\text{rel}}} = \frac{150\text{ m}}{15\text{ m/s}} = 10\text{ seconds}$.
Correct Answer: (A)
Solution 3:
In relative frame of the bus:
Initial relative velocity of man: $u_{\text{rel}} = v – 0 = v$.
Relative acceleration: $a_{\text{rel}} = a_{\text{man}} – a_{\text{bus}} = 0 – 1.0 = -1.0\text{ m/s}^2$.
For the man to catch the bus, the distance covered relative to bus before relative speed drops to zero must be at least $48\text{ m}$:
$v_{\text{rel}}^2 = u_{\text{rel}}^2 + 2 a_{\text{rel}} s \implies 0 = v^2 – 2(1.0)(48) \implies v^2 \ge 96 \implies v \ge \sqrt{96} = 4\sqrt{6}\text{ m/s} \approx 9.8\text{ m/s}$.
Correct Answer: (D)
Solution 4:
Speed of police van: $v_p = 30\text{ km/h} = 30 \times \frac{5}{18} = \frac{25}{3}\text{ m/s}$.
Muzzle speed of bullet relative to van: $v_{b/p} = 150\text{ m/s}$.
Speed of bullet in ground frame: $v_b = v_{b/p} + v_p = 150 + \frac{25}{3} = \frac{475}{3}\text{ m/s}$.
Speed of thief’s car: $v_t = 192\text{ km/h} = 192 \times \frac{5}{18} = \frac{160}{3}\text{ m/s}$.
Relative speed with which bullet hits thief’s car:
$v_{\text{hit}} = v_b – v_t = \frac{475}{3} – \frac{160}{3} = \frac{315}{3} = 105\text{ m/s}$.
Correct Answer: (A)
Solution 5:
For two dropped bodies, $u_1 = u_2 = 0$ and $a_1 = a_2 = g$.
Relative acceleration: $a_{\text{rel}} = g – g = 0$.
Relative velocity: $v_{\text{rel}} = 0 – 0 = 0$.
Relative displacement: $\Delta y(t) = \Delta y(0) + v_{\text{rel}} t + \frac{1}{2}a_{\text{rel}} t^2 = (h_1 – h_2) + 0 + 0 = h_1 – h_2 = \text{constant}$.
Correct Answer: (C)
Solution 6:
– (A) True: $v_B – v_A = (u_B – gt) – (u_A – gt) = u_B – u_A = 30 – 20 = 10\text{ m/s}$.
– (B) True: $a_{\text{rel}} = (-g) – (-g) = 0$.
– (C) True: $\Delta y = v_{\text{rel}} \cdot t = 10t$.
– (D) $A$ reaches highest point at $t_A = \frac{u_A}{g} = \frac{20}{10} = 2\text{ s}$. At $t = 2\text{ s}$, separation is $\Delta y = 10(2) = 20\text{ m}$. (True)
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– In the train frame, horizontal velocity is zero, so the path is a vertical straight line (A is True).
– In the ground frame, horizontal velocity is $v$ and vertical is $u – gt$, tracing a parabola (B is True).
– Vertical motion is identical ($T = \frac{2u}{g}$), so time of flight and maximum height are frame-independent (C, D are True).
All statements are correct.
Correct Answer: (A, B, C, D)
Solution 8:
(A), (B), and (C) correctly express the mathematical criteria for collision avoidance in relative coordinates.
(D) is false: if $a_A < a_B$, the car decelerates slower than the truck, meaning the relative speed $u_A - u_B$ increases or does not diminish fast enough, making collision unavoidable if $d$ is insufficiently large.
Correct Answer: (A, B, C)
Solution 9:
$v_A = 54 \times \frac{5}{18} = 15\text{ m/s}$.
$v_B = 36 \times \frac{5}{18} = 10\text{ m/s}$.
$v_{\text{rel}} = 15 + 10 = 25\text{ m/s}$.
$s_{\text{rel}} = L_A + L_B = 100 + 100 = 200\text{ m}$.
$T = \frac{200\text{ m}}{25\text{ m/s}} = 8\text{ seconds}$.
Correct Answer: 8
Solution 10:
$t = \frac{t_1 t_2}{t_1 + t_2} = \frac{90 \times 60}{90 + 60} = \frac{5400}{150} = 36\text{ seconds}$.
Correct Answer: 36