COMEDK Exam 2022 Solved Paper

Given, speed of car, u=54km∕h=54×

5

18

=15m∕s
Since, car stops in 3 s on applying brakes, so acceleration can be calculated from first equation of motion,
v=u+at⇒0=15+a(3)
⇒ a=−

15

3

=−5ms−2
Using second equation of motion,
s=ut+

1

2

at2
=15×3+

1

2

×(−5)×(3)2
=45−22.5=22.5m