A tortoise challenged a hare to a 100-meter race on a track

Question

A tortoise challenged a hare to a 100-meter race on a track. The tortoise negotiated a 50-meter head start with the hare. When the race started the hare was running at a constant speed of 2.8 meters per second and the tortoise was crawling at a speed of 0.5 meters per second. (They both maintained these speeds for the entire race.)

 

Our goal is to determine who won the race.

1. Read the above problem statement again, then explain in writing on a sheet of paper how you will determine who won the race.
2. Construct a drawing to represent the 100-meter length of the track. Then place the tortoise and hare's starting points on the track.
3. Define the variable t to represent the number of seconds since the start of the race.
4. Write an expression to represent the Tortoise's distance from the starting line in terms of t.  
5. Represent the Hare's distance from the starting line in terms of t.     
6. Write a formula to represent the distance, d (in meters), that the tortoise is ahead of the hare in terms of t, the amount of time since the start of the race.
7. What is the value of t when the Hare catches up to the Tortoise? 
8. How far has the hare run from the start of the race when he catches up to the tortoise?    
9. On the graph below:
10. construct a graph that represents the distance, dd, that the tortoise is ahead of the hare in terms of the number of seconds, tt, since the start of the race.
11. Plot the point (10,27)(10,27) on your graph.

Answer 1

hare: 2.8 = speed, 100= distance, d=rate of speed times t

time=d/r = 100/2.8 = about 35.7 seconds to reach the finish line

tortoise: .5 = r

time = 50/.5 = 100 seconds to finish

hare wins

distance from starting line, d = rt

then

for hare: d=2.8t

for tortoise d=50+.5t

for distance tortoise is ahead dd=50+.5t - 2.8t = 50-2.3t

dd=50-2.3t for t=0 to t=50/2.3 = about 21.7 seconds when the hare catches up

for t>21.7 d=50-2.3t will be negative

hare runs d=2.8(50/2.3) = about 60.87 meters when it catches the tortoise

dd=50-2.3t graphically, with t on the horizontal or x axis and dd on the vertical or y axis is a straight line with y or dd intercept of 50 and slope =-2.3

(10,27) is when the tortoise is 27 meters ahead after 10 seconds