Question

Carol and Harry rode go carts around a National park service path. Carol rode 12 miles per hour. Harry left 15 minutes later and rode 16 miles per hour. After how many miles did Harry catch up to to Carol?

Answer 1

The distance travelled by Carol is given by the product between the time she travelled and the rate she rides. The same logic is applied to Harry.

$\begin{gathered} d_c=12\cdot t_c \\ d_h=16\cdot t_h \end{gathered}$

If we use our time variables using hour as the unit, let's start by converting the time difference between their departure to hours.

$\begin{gathered} 1h=60\min \\ \frac{1h}{4}=\frac{60\min }{4} \\ 0.25h=15\min \end{gathered}$

The difference between their departure is 0.25h. Since Harry departed later, the total time travelled by Carol when Harry catch up is the time Harry travelled plus 0.25h.

$t_c=t_h+0.25$

Renaming the time travelled by Harry as t, our initial equations turns out to be

$\begin{gathered} d_c=12\cdot(t+0.25) \\ d_h=16\cdot t \end{gathered}$

When Harry catch up, the distance travelled by them will be the same, therefore

$12\cdot(t+0.25)=16\cdot t$

Solving for t, we have

$\begin{gathered} 12\cdot(t+0.25)=16\cdot t \\ 12t+3=16t \\ 3=4t \\ t=\frac{3}{4} \\ t=0.75 \end{gathered}$

Now we know that Harry catch up with carol after 0.75h(45 minutes). Plugging this value in our expression, we have

$d_h=16\cdot0.75=12$

Harry travelled 12 miles to catch up with Carol.