# College Physics| Openstax| Problem 2.39

#### SOLUTION:

There are two parts to the race: an acceleration part and a constant speed part. First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, $a=\frac{600\:mph}{4\:s}$. $v_{max}=at_1$ $t_1=\frac{v_{max}}{a}$ $t_1=\frac{183\:mph}{\left(\frac{60\:mph\:}{4\:s}\right)}$ $t_1=12.2\:s$ $x_1=v_{avg}t_1$ $x_1=\left(\frac{183\:mph+0\:mph}{2}\right)\left(12.2\:s\right)$ $x_1=\left(91.5\:mph\right)\left(\frac{1\:h}{3600\:s}\right)\left(12.2\:s\right)$ $x_1=0.31\:mi$

At constant velocity, $x_2=5\:mi-0.31\:mi=4.7\:mi$.

Now, we complete the calculation by determining how much time is spent on the course at max speed. $t_2=\frac{x_2}{v_{max}}$ $t_2=\frac{4.7\:mi}{183\:mph}$ $t_2=\left(0.026\:h\right)\left(\frac{3600\:s}{1\:h}\right)$ $t_2=92\:s$

So, the total time is $t_{total}=t_1+t_2$ $t_{total}=12.2\:s+\:92\:s$ $t_{total}=104\:s$

You can now buy the complete solution manual of College Physics by Openstax. Just click on the button you can see at the right portion of this post.

Advertisements