The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m. i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?

i) a = 0.977 [m/s²]

ii) x = 115.06 [m]

Explanation:

In order to determine the acceleration, we must use the following equation of kinematics.

where:

Vf = final velocity = 7 [m/s]

Vo = initial velocity = 15 [m/s]

a = acceleration [m/s²]

x = displaciment = 90 [m]

Now replacing:

When the bus coming to rest.