Question #165270

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


1
Expert's answer
2021-02-22T10:23:12-0500

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


vi=xit=8900 m27 min60 s1 min=5.5 ms.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 ms2.0.2\ \dfrac{m}{s^2}. We can write the velocity function of the runner as follows:


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

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


d1(t)=0tv(t)dt,d_1(t)=\displaystyle\intop_{0}^tv(t)dt,d1(t)=0t(5.5+0.2t)dt=5.5t+0.1t2.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 (180t)(180-t) seconds at constant speed (after he accelerates tt seconds):


x2(t)=(5.5+0.2t)(180t).x_2(t)=(5.5+0.2t)(180-t).

Finally, adding these two distances, we get:


xfxi=d1(t)+d2(t),x_f-x_i=d_1(t)+d_2(t),5.5t+0.1t2+(5.5+0.2t)(180t)=100008900=1100,5.5t+0.1t^2+(5.5+0.2t)(180-t)=10000-8900=1100,0.1t236t+110=0.0.1t^2-36t+110=0.

This quadratic equation has two roots: t1=357 st_1=357\ s and t2=3.1 s.t_2=3.1\ s. Since the runner has only 180 seconds to reach the destination, the correct answer is t=3.1 s.t=3.1\ s.

Answer:

t=3.1 st=3.1\ s.


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