Answer to Question #171115 in Physics for DAnte

Question

at the end of the race, a runner decelerates from a velocity of 9m/s^2. how far does she travel in next 5s?

Answer 1

I assume, the initial velocity was $v_0 = 9m/s$. Then the acceleration was (by definition):

$$a = \dfrac{v_f-v_0}{t} = -\dfrac{v_0}{t}$$

where $v_f = 0$, is the final speed of the runner, $t = 5s$ is the time she decelerated.

The distance travelled in this time was:

$$d = v_0t - \dfrac{at^2}{2} = v_0t - \dfrac{v_0t}{2} = \dfrac{v_0t}{2}$$

Thus, obtain:

$$d = \dfrac{9m/s\cdot 5s}{2} = 22.5m$$

Answer. 22.5 m.