Answer to Question #165270 in Physics for Jaja Cals

Question

A runner wants to run 10,000m in 30 minutes. After 27 minutes he has run 8,900m. For how long must he accelerates at 0.02 m/s² and then maintain that final speed

Answer 1

Let's first find the initial velocity of the runner (before he begins to accelerate):

$$v_i=\dfrac{x_i}{t}=\dfrac{8900\ m}{27\ min\cdot\dfrac{60\ s}{1\ min}}=5.5\ \dfrac{m}{s}.$$

Then, the runner begins to accelerate at a constant acceleration $0.2\ \dfrac{m}{s^2}.$ We can write the velocity function of the runner as follows:

$$v(t)=5.5+0.2t$$

We can find the distance at which the runner will accelerate by integrating $v(t)$ from 0 to $t$:

$$d_1(t)=\displaystyle\intop_{0}^tv(t)dt,$$ $$d_1(t)=\displaystyle\intop_{0}^t(5.5+0.2t)dt=5.5t+0.1t^2.$$

Then, we can find the distance that the runner will run in $(180-t)$ seconds at constant speed (after he accelerates $t$ seconds):

$$x_2(t)=(5.5+0.2t)(180-t).$$

Finally, adding these two distances, we get:

$$x_f-x_i=d_1(t)+d_2(t),$$ $$5.5t+0.1t^2+(5.5+0.2t)(180-t)=10000-8900=1100,$$ $$0.1t^2-36t+110=0.$$

This quadratic equation has two roots: $t_1=357\ s$ and $t_2=3.1\ s.$ Since the runner has only 180 seconds to reach the destination, the correct answer is $t=3.1\ s.$

Answer:

$t=3.1\ s$.