First slide

Rectilinear Motion

Question

A man is" d" distance behind the bus. when the bus starts from rest with an acceleration $a_{0}$ , With what minimum constant velocity should the man start running to catch the bus

Moderate

A

$\sqrt{4 a_{0} d}$
B

$\sqrt{3 a_{0} d}$
C

$\sqrt{2 a_{0} d}$
D

$\sqrt{a_{0} d}$

Solution

$c o n s \tan t v e l o c i t y o f m a n b e v l e t T h e m a n c a t c h e s t h e b u s a t p o i n t P, a f t e r t i m e t f o r U n i f o r m m o t i o n : \to d + s = V t from figure s = v t - d -- \to (1)$
$b u s a c c e l e r a t e f r o m r e s t, a c c e l e r a t i o n o f b u s i s a_{0} s = 0 + \frac{1}{2} a_{0} t^{2} \to (2) s u b s t i t u t e e q u a t i o n (1) v t - d = \frac{1}{2} a_{0} t^{2} \Rightarrow a_{0} t^{2} - 2 v t + 2 d = 0 T h e m a n w i l l c a t c h t h e b u s, i f t i m e t i s r e a l, f o r t i m e t t o b e c o n s i d e r d i s c r i m i n a n t$
   $[{(- 2 V)}^{2} - 4 a_{0} \cdot 2 d] \geq 0$
   $V^{2} - 2 a_{0} d \geq 0$
$V \geq \sqrt{2 a_{0} d}$
   $V_{\min} = \sqrt{2 a_{0} d}$

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