Question

A driver of a car travelling at 52km/h applies the breaks and accelerates uniformly in the opposite direction. the car stops in 5s. another driver going at 37kn/h in another car applies his breaks slowly and stops in 10s.which of the two cars travelled farther after applying breaks​

Answer 1

Acceleration of first car =

U = 52km/h = 14.44 m/s.

V=0 (because car stops)

t = 5 seconds.

$by \: formula \: v = u + at \\ \\ 0 = 14.44 + 5a \\ - 14.44 \div 5 = a \\ = - 2.88 {ms}^{ - 2} \\ \\ distance \: travelled \: is \:$

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ 14.44 \times 5 + \frac{1}{2} ( - 2.88) \times 25 \\ 72.2 - 36 \\ = 36.2m$

distance by second car

37 km/h= 10.277m/s

by formula v=u+at

0= 10.27+a10

a=-1.027 ms-²

distance traveled

s=ut+ 1/2 at²

10.27x10+(-1.027x50)

102-51.38

=50.62m

so car B travelled more.