Question

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brakes and slows to 8.0 km/hr in 0.25 sec. What is the acceleration of the bike rider during this period?

Answer 1

Step-by-step explanation:

First we have to change the velocity to km/s since the time is given in seconds.

12.0 km/hr = 3.33 ×
`10^{-3` and

8.0 km/hr = 2.2 ×
`10^{-3`

The equation for acceleration is

`a=(v_f-v_0)/(t)` and filling in:

`a=(2.2*10^(-3)-3.33*10^(-3))/(.25)=(-.00113)/(.25)=-.0045(km)/(s^2)`

Answer 2

Acceleration of the bike rider during this period is -16 km/hr/s