College Physics by Openstax Chapter 2 Problem 25

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: v0=9.00 m/sv_0=9.00 \ \text{m/s}; and a=2.00 m/s2a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 st=5.00 \ \text{s} and we shall use the formula x=x0+v0t+12at2 x=x_0+v_0 t+\frac{1}{2}at^2.

x=x0+v0t+12at2x=0 m+(9.00 m/s)(5.00 s)+12(−2.00 m/s2)(5.00 s)2x=20 meters  (Answer)\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 vf=v0+atv_f=v_0+at.

vf=v0+atvf=9.00 m/s+(−2.00 m/s2)(5.00 s)vf=−1 m/s  (Answer)\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.

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