Question

A car goes from 90.0m/s to a stop in 3.00s. What is the acceleration

Answer 1

`-30m/s^(2)`

Step-by-step explanation:

The first equation of motion given as,

v = u +at

Here, u is initial and v is final velocity and a is acceleration and t is time taken.

As per question car initial velocity, u = 90 m/s, final velocity, v =0 and time taken, t = 3.00 s.

Substituting these values in first equation of motion, we get

0 = 90 m/s + a×3.00 s

or,
`a = (-90m/s)/(3.00s)`

`a=-30m/s^(2)`

Thus, the deacceleration of the car is
`-30m/s^(2)`