A tortoise and a hare are having a 100 m race. When the starting gun goes off, t

Question

A tortoise and a hare are having a 100 m race. When the starting gun goes off, the hare lies down for a nap, while the tortoise moves forward with a constant acceleration, reaching a speed of 2.0 m/s when she is 20 m from the starting line. After this, the tortoise travels at a constant velocity
of 2.0 m/s until crossing the finish line. After 45 seconds the hare wakes up from his nap, and covers the 100 m with a constant acceleration of 2.0 m/s2.
(a) 50 seconds after the starting gun is fired, how far from the starting line is the
tortoise? How far from the starting line is the hare?
(b) Which animal wins the race? Clearly justify your answer.
(c) How much time passes between the winner reaching the finish line and the other
animal reaching the finish line?
(d) What is the distance between the animals when the winner crosses the finish line?
(e) What is the distance between the animals at the only time (other than at the instant the starting gun is fired) they have the same velocity?

Explanation / Answer

Let's get an equation for the distance the tortoise travels with respect to time. We will call this xT(t).

For xT(t) < 20 m. Let's say the tortoise reaches the 20 m mark at t = t20

xT(t) = (1/2)at2
vT(t) = at
xT(t20) = 20 = (1/2)at202
vT(t20) = 2 = at20

So,
a = 2/t20
Since,
20 = (1/2)at202
40 = at202
40 = (2/t20)t202 = 2t20
t20 = 20 s
a = 0.1 m/s2

So, for t < 20 seconds,
xT(t) = 0.05t2
vT(t) = 0.1t

for t >= 20 seconds,

vT(t) = 2.0 m/s
Since the acceleration is 0 because of constant velocity, then it's just a distance = rate*time formula for the distance:
xT(t) = 2.0*(t-20)

As for the hare,
For t < 45 seconds,
xH(t) = vH(t) = aH(t) = 0

For t > 45 seconds,
aH(t) = 2.0 m/s2
vH(t) = 2(t-45)
xH(t) = (t-45)2

(a) t = 50 seconds
Tortoise:
In the first 20 seconds, the tortoise goes 20 meters.
For the remaining 30 seconds,
xT(30) = 2.0*(t - 20) = 2.0*10 = 20 m
So, the total distance is 40 meters for the tortoise.
The hare travels:
xH(50) = (50-45)2 = 52 = 25 m
The hare travels only 25 meters.

(b) The race is won by whomever reaches the 100 m mark the fastest.
For the tortoise, after it accelerates through the first 20 m in 20 s, it needs to travel 80 m. Let's see how long that takes.
xT(t) = 80 = 2.0*(t-20)
40 = t - 20
60 = t
So, it takes an additional 60 seconds, for a grand total of 80 seconds to finish.
For the hare,
xH(t) = 100 = (t-45)2
10 = t - 45
t = 55
So, after the hare wakes up after sleeping for 45 seconds, it only takes it 10 seconds to finish, for a grand total of 55 seconds, making it the first to reach the finish line from when the starting gun was fired.

(c) From our calculation above, only 5 seconds pass between the hare's finish and the tortoise's.

(d) The winner crosses the finish line at the 55 second mark. Let's see where the tortoise is at that time.
xT(55) = 2.0*(55-20) = 2*35 = 70 m
Add this to the 20 m it covers before traveling at a constant velocity and we get 90 m. So there were 10 m of distance between the tortoise and the hare at the end.

(e) So, the tortoise and hare have the same velocity after the hare wakes up, which means the tortoise was traveling at constant velocity. Let's set these equations equal:
vT(t) = vH(t)
2 = 2(t-45)
1 = t - 45
t = 46
So, at the 46 second mark, the velocities were equal. Let's find their distances at this time.
xT(46) = 2.0*(46-20) = 2*(26) = 52 m
And then we add the 20 m it traveled during the acceleration portion, to give 72 m.
As for the hare,
xH(46) = (46-45)2 = 1 m
The hare only traveled one meter at that time. The distance between the animals is 71 m.

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