Question

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?

Answer 1

Answer:

(a) 2.29 km/h

(b) 9 km/h

Step-by-step explanation:

For part (a) you have to apply the average speed formula, which is defined by:

$v=\frac{d}{t}$

where d is the total distance traveled and t is the total time needed.

$d=\frac{4.00}{1.75}=2.29$ km/h

For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:

The nap time in hours is:

90/60 = 1.5 h (because there are 60 minutes in one hour)

The running time is:

T= 1.75 - 1.5 = 0.25 h

Let t1 represent the time before the nap and t2 the time after the nap:

t1+t2 = T

t1+t2 = 0.25

You have to apply the formula d=vt before and after the nap:

-Before the nap, the distance traveled was 0.50 km

0.50 = v1t1

-Afer the nap, the distance traveled was 3.50 km

3.50=v2t2

But v2=2v1 (because after the nap the rabbit runs twice as fast)

You have to solve the system of equations:

t1=0.25-t2 (I)

v1t1=0.50 (II)

2v1t2=3.50 (III)

Replacing (I) in (II)

v1(0.25-t2)=0.50

Applying distributive property and solving:

0.25v1-v1t2=.050

For (III) you have that v1t2=3.50/2=1.75. Hence:

0.25v1-1.75=0.50

Solving for v1:

v1 = 9 km/h