At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s².

(a) How far does she travel in the next 5.00 s?

(b) What is her final velocity?

(c) Evaluate the result. Does it make sense?

Solution:

We are given the following: v_0=9.00 \ \text{m/s}; and a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 \ \text{s} and we shall use the formula  x=x_0+v_0 t+\frac{1}{2}at^2.

\begin{align*}
x & =x_0+v_0 t+\frac{1}{2}at^2 \\
x & =0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2 \\
x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

The final velocity can be determined using the formula v_f=v_0+at.

\begin{align*}
v_f & =v_0+at \\
v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\
v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s². After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.