Question

a motorcycle was traveling at a speed of 32 m/s. it took the motorcycle a distance of 75 m to stop. what was the motorcycles rate of deceleration.?

Answer 1

Deceleration value of motorcycle =
`6.83 m/s^2`

Explanation:

We have equation of motion,
`v^2=u^2+2as`, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

In this case we have final velocity = 0 m/s, initial velocity = 32 m/s, and displacement = 75 m, now we need to find deceleration value.

Substituting

`0^2 = 32^2+2*a*75\\ \\ a = -6.83 m/s^2`

So, Acceleration value of motorcycle =
`-6.83 m/s^2`

Deceleration value of motorcycle =
`6.83 m/s^2`