Question 2.1: THE TORTOISE AND THE HARE GOAL Apply the concept of average ...
THE TORTOISE AND THE HARE
GOAL Apply the concept of average speed.
PROBLEM A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0 – min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75 h, the rabbit wins the race. (a) Calculate the average speed of the rabbit. (b) What was his average speed before he stopped for a nap? Assume no detours or doubling back.
STRATEGY Finding the overall average speed in part (a) is just a matter of dividing the path length by the elapsed time. Part (b) requires two equations and two unknowns, the latter turning out to be the two different average speeds: v 1 before the nap and v_2 after the nap. One equation is given in the statement of the problem (v_2 = 2v_1), whereas the other comes from the fact the rabbit ran for only 15 minutes because he napped for 90 minutes.
(a) Find the rabbit’s overall average speed.
Apply the equation for average speed:
\begin{aligned}\text { Average speed } & \equiv \frac{\text { path length }}{\text { elapsed time }}=\frac{4.00 \mathrm{~km}}{1.75 \mathrm{~h}} \\&=2.29 \mathrm{~km} / \mathrm{h}\end{aligned}
(b) Find the rabbit’s average speed before his nap.
Sum the running times, and set the sum equal to 0.25 \mathrm{~h} :
t_{1}+t_{2}=0.25 \mathrm{~h}
Substitute t_{1}=d_{1} / v_{1} and t_{2}=d_{2} / v_{2} :
(1) \frac{d_{1}}{v_{1}}+\frac{d_{2}}{v_{2}}=0.25 \mathrm{~h}
Substitute v_{2}=2 v_{1} and the values of d_{1}[latex] and [latex]d_{2} into Equation (1):
(2) \frac{0.500 \mathrm{~km}}{v_{1}}+\frac{3.50 \mathrm{~km}}{2 v_{1}}=0.25 \mathrm{~h}
Solve Equation (2) for v_{1} :
v_{1}=9.0 \mathrm{~km} / \mathrm{h}
REMARKS As seen in this example, average speed can be calculated regardless of any variation in speed over the given time interval.