Question

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?

Answer 1

i) a = 0.977 [m/s²]

ii) x = 115.06 [m]

Step-by-step explanation:

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

`v_(f) ^(2) =v_(o) ^(2) -2*a*x`

where:

Vf = final velocity = 7 [m/s]

Vo = initial velocity = 15 [m/s]

a = acceleration [m/s²]

x = displaciment = 90 [m]

Now replacing:

`(7)^(2) =(15)^(2) -2*a*90\\2*a*90 = 15^(2) - 7^(2) \\180*a=176\\a=0.977[m/s^(2)]`

When the bus coming to rest.

`v_(f)^(2) =v_(o) ^(2) -2*a*x\\0 = 15^(2) -2*0.977*x\\x = 115.06[m]`