Question

A car moving with a velocity of $20m{s}^{-1}$ is stopped in a distance of $40m$. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

• A. 640 m
• B. 320 m
• C. 1280 m
• D. 160 m

Answer 1

Answer: (D)

Initial velocity of car, $u=20m/s$
The final velocity will be, $v=0$ as the car comes to rest after travelling distance of $40m$ (which will be after application of brakes)
By third equation of motion we know,
$<br/>\begin{array}{l}<br/>v^2=u^2+2\text{ as }\\ <br/>0=(20{)}^2+2\times a\times (40)\\ <br/>-80a=400<br/>\end{array}<br/>$
Thus, $a=-5m/s^2$
Negative sign indicates that it is a retardation.
When same car is moving with double the velocity
Initial velocity, $u=2\times 20=40m/s$
Thus,
$<br/>\begin{array}{l}<br/>0=(40{)}^2\times 2\times (-5)\times s\\ <br/>10s=1600\\ <br/>s=160m<br/>\end{array}<br/>$
Hence, if the same car is travelling at double the velocity the distance travelled by it for same retardation is $160m$