Question

A car moving along a straight highway with a speed of $126km/hr$ is brought to a stop within a distance of $200m$ . What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Answer 1

Step 1. Given data:

$$\begin{gathered} Initialvelocity\left(u\right)=126Km{h}^{-1} \\ Initialvelocity\left(u\right)=\frac{126}{3600}\times 1000 \\ Initialvelocity\left(u\right)=35m{s}^{-1} \end{gathered}$$

$Dis\tan ce\left(s\right)=200m$

$Finalvelocity\left(v\right)=0$, since the car will stop after some time

Step 2. Formula used:

3rd Equation of Motion

$v^2=u^2+2as$

Where $v$ is the final velocity,

$u$ is initial velocity,

$a$ is acceleration and

$s$ is distance

First Equation of Motion

$v=u+at$

Where $t$ is the time taken

Step 3. Calculating retardation of the car,
$v^2=u^2+2as$

$a=-\frac{u^2}{2s}$

$a=\frac{-u^2}{2s}$

$=-3.06\frac{m}{s^2}$

The negative sign indicates that the car is retarding.

Retardation is nothing but negative acceleration.

Step 4. Calculating time taken,

$v=u+at$

$t=\frac{-u}{a}$

$=\frac{-35}{-3.06}$

$t=11.43s$

Hence, the retardation of car is $-3.06m{s}^{-2}$ and time taken by the car to stop is $t=11.43s$