First slide

Rectilinear Motion

Question

When the speed of a car is V, the minimum distance over which it can be stopped is x. If the speed becomes ‘nv’, what will be minimum distance over which it can be stopped during same time.

Moderate

A

$' n x'$
B

$\frac{x}{n}$
C

$n^{2} x$
D

$\frac{x}{n^{2}}$

Solution

From Kinematic equation of motion $V^{2} = u^{2} + 2 a S$
Given $S = x$ , $V = 0$
Maximum retardation, $a = \frac{V^{2}}{2 S} = \frac{V^{2}}{2 x}$
When the initial velocity is nv, then the distance over which it can be stopped is given by $S^{1} = \frac{u^{1^{2}}}{2 a} (u^{1} = n v)$
$\begin{matrix} S = \frac{(n v)^{2}}{2 (\frac{V^{2}}{2 x})} = \frac{n^{2} V^{2}}{(\frac{V^{2}}{x})} \\ = n^{2} x \end{matrix}$

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