First slide

Relative motion based problems

Question

A body is moving with uniform velocity of 8 ${ms}^{- 1}$ . When the body just crossed another body, the second one starts and moves with uniform acceleration of $4 {ms}^{- 2}$ . The distance covered by the second body when they meet is:

Moderate

A

8 m
B

16 m
C

24 m
D

32 m

Solution

$S_{r e l} = U_{r e l} t + \frac{1}{2} a_{r e l} t^{2}$
$0 = - 8 t + \frac{1}{2} 4 t^{2}$
$t = 2 s$
distance covered before meeting
s = 8t = 8 $\times$ 4 = 32 m

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