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

Answer:

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

$a=\frac{v_f-v_0}{t}$ and filling in:

$a=\frac{2.2*10^{-3}-3.33*10^{-3}}{.25}=\frac{-.00113}{.25}=-.0045\frac{km}{s^2}$

Answer 2

Answer:

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