First slide

Relative motion based problems

Question

The velocity-time graph of robber's car and a chasing police car are shown in the following graph. Police car crosses the robber's car in time

Moderate

A

10 s after it starts
B

15 s after it starts
C

20 s after it starts
D

Never crosses

Solution

From graph, velocity of robber's car = $10 {ms}^{- 1}$
Let police car crosses it after t second.
Distance travelled by robber's car = 10 t m
Police car is moving with a constant acceleration of $1 {ms}^{- 2}$ as it attains a velocity of $10 {ms}^{- 1}$ in 10 s after starting from rest. Distance travelled in t second = $\frac{1}{2} \cdot a \cdot t^{2} = \frac{1}{2} t^{2}$
When the police car crosses the robber's car, distance travelled by the both cars should be same from the starting
point of chase.
$∴ \frac{1}{2} t^{2} = 10 t \Rightarrow t = 20 s$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App