Question

The brakes applied to a car it produce an acceleration of -10 m/s2. If the car takes 5s to stop after applying the brakes,
calculate the distance covered by the car before coming to rest.​

Answer 1

Answer:

125m

Explanation:

here we have :

final velocity, v : 0 ( as the car comes to rest)

acceleration, a: -10 m s^-1

time ,t : 5 secs

let initial velocity be u and,

distance covered be s

now by first equation of motion,

v = u + at

therefore, 0 = u + (-10)*5

u = 50 ms^-1

now, by second equation of motion,

s = ut + 1/2 at^2

= 50*5 + 1/2* (-10)*5^2

= 250 - 125

= 125

therefore,. distance,s = 125m

The car covers 125m before coming to rest.

HOPE IT HELPS :)

Answer 2

Answer:

125 m

Explanation:

Given,

          Final Velocity, v + 0 m/s (the car comes to rest)

          Acceleration, a = -10 m/s²

          Time Takes, t = 5 s

Let Initial Velocity be 'u' and the distance covered 's'

Using the first Equation of Motion-

                     v = u+at

                     0 = u + (-10) × 5

                     0 = u + (-50)

                     u = 50 m/s

Now,

        using the second Equation of Motion-

                      s = ut+(1/2)at²

                      s = 50 × 5 + (1/2) × (-10) × 25

                      s = 250 - 125

                      s = 125m

Therefore, the car covered 125 m before coming to rest