Question

By applying the brakes without causing the skid the driver of a car is able to stop his car within a distance of 5 metre if it is going at 36 kmph, if the car were going at 72 kmph using the same brakes he can stop the car over a distance of​
a- 10 m
b- 2.5 m
c- 20 m
d- 40 m

Answer 1

Hint: We are given initially the car was moving and it stops, so acceleration would be negative called retardation. In the second case the speed is doubled, we can make use of Newton’s equations of motion to solve this problem. The brakes remain the same, so the value of deceleration will be the same.

Complete step by step answer:Let us represent initial velocity by u, final velocity by v and distance by s.
u= 36 km/h \[=36\times \dfrac{5}{18}=10m/s\]
v= 0
s= 5 m
using,
$
{{v}^{2}}-{{u}^{2}}=2as \\
\implies 0-100=2a\times 5 \\
\implies a=-10m{{s}^{-2}} \\
$
Now in the second case u= 72km/h,
\[=72\times \dfrac{5}{18}=20m/s\], now again using second equation of motion we get,
$
{{v}^{2}}-{{u}^{2}}=2as \\
\implies 0-400=2\times -10\times s \\
\implies s=20m \\
$

So, the correct answer is C

Note- since acceleration is defined as the rate of change of velocity and velocity itself is a vector quantity, so, acceleration is also a vector quantity, so it can assume negative values also. A positive value of acceleration means the speed of the body is increasing while the negative value of acceleration means the velocity of the body is decreasing. Negative acceleration also termed as retardation. If a body moves with constant velocity then acceleration produced in the body is zero.