Question

the brakes of a car provide a maximum retardation of 10ms^-2.The least distance in which the car can be stopped if its velocity is 30ms/s will be​

Answer 1

Answer:

a= -10m/s²

v=0

u=30m/s

t=?

a= v-u/t

t= v-u/a

t= o-30/-10

t= 3sec

Answer 2

Answer:

45m

Explanation:

Given:

a = -10m/s

v = 0

u = 30m/s

To find: Least distance in which car can be stopped (s).

Solution:

We know that,

v² = u² + 2as

(0)² = (30)² + 2 * (-10) * s

0 = 900 - 20s

-900 = -20s

900 = 20s

s = 900/20

s = 45m

Answer: Least distance in which car can be stopped (s) = 45m