Question

If the car's speed decreases at a constant rate from 77 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line?

Answer 1

The acceleration of the car,

`a = (v-u)/(t)`

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given
`u =77 \ mil/h` ,
`v = 50 mi/h` and
`t = 3.0 s = 3.0 * (1 \ h)/(3600) = 8.3 * 10^(-4) h`

Therefore, from above equation

`a =  (50 \ mi/h -77 \ mi/h)/(8.33 * 10^(-4) h) = - (27 \ mi/h )/(8.33 * 10^(-4) h) = - 3.2 * 10^(4) \ mi/h^2`.

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is
`3.2 * 10^(4) \ mi/h^2`.