Question

A car travelling at 15 m/s comes to rest due to application of brakers that produces a deceleration of 5 m/ S2 . the shoping distance is​

Answer 1

Answer:

Using,

v = u + at

v = 0

u = 15 m/s

a = - 5 m/s^2

0 = 15 - 5*t

t = 15/5 = 3 d

Now, using

s = ut + 1/2 a t^2

= 15*3 + 1/2 * (-5) * 3*3

= 45 - 45/2

= (90 - 45)/2

= 45/2 = 22.5 m

Explanation:

i this is right ☺️