Question

A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a distance of

75.0 m. If the vehicle's acceleration remained constant during the braking, what was the vehicle's initial speed?

Answer 1

Since acceleration is constant, the average and instantaneous accelerations are the same, so that

`a = a_(\rm ave) = (\Delta v)/(\Delta t) = -(v_i)/(10.0\,\rm s)`

By the same token, we have the kinematic relation

`v^2 - {v_i}^2 = 2a\Delta x`

where
`v` is final speed,
`v_i` is initial speed,
`a` is acceleration, and
`\Delta x` is displacement.

Substitute everything you know and solve for
`v_i` :

`0^2 - {v_i}^2 = 2\left(-(v_i)/(10.0\,\rm s)\right)(75.0\,\mathrm m)`

`\implies {v_i}^2 - \left(15.0(\rm m)/(\rm s)\right) v_i = 0`

`\implies v_i \left(v_i - 15.0(\rm m)/(\rm s)\right) = 0`

`\implies v_i = \boxed{15.0(\rm m)/(\rm s)}`