Question

A car is travelling with a constant speed when driver applies break giving deceleration of 3.50 m/s2.If the car stoops at distance of 30m.What was cars original speed?.

Answer 1

Answer:

acceleration= -3.5m/s^2

distance travel=s=30 m

let speed of the car be x

final velocity=0

using

2as=v^2-u^2

2×(-3.5)×30= 0-x^2

210= x^2

x=√210

x= 14.5

hece original speed of the car is 14.5 m/s

hope this will help you