Question

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 0.8 m/s2 for 4.4 seconds. It then continues at a constant speed for 13.5 seconds, before getting tired and slowing down with constant acceleration coming to rest 61.0 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop.
1.How fast is the hare going 2.2 seconds after it starts?
2. How fast is the hare going 15.1 seconds after it starts?
3. How far does the hare travel before it begins to slow down?
4. What is the acceleration of the hare once it begins to slow down?
5. What is the total time the hare is moving?
6. What is the acceleration of the tortoise?

Answer 1

Answer:

Question 1

The velocity of the hare 2.2 s after it starts is 1.76 m/s

Question 2

15.1 s after the hare starts, its velocity is 3.53 m/s

Question 3

The hare travels 55.49 m before it begins to slow down.

Question 4

Once it begins to slow down, the acceleration of the hare is -1.13 m/s²

Question 5

The total time the hare is moving is 21.02 s.

Question 6

The acceleration of the tortoise is 0.28 m/s².

Explanation:

These kinematic equations apply for the hare:

for the first 4.4 seconds:

v = v0 + at

x = x0 + v0t + 1/2at²

where:

v = velocity

v0 = initial velocity

a = acceleration

t = time

x = position

x0 = initial position

from 4.4 s to 17.9 s (+13.5 s)

v = constant.

the velocity is the same as the final velocity in the first 4.4 s of the race:

v = v0 + a*4.4s

x = x0 + vt

from 17.9 s until end:

v = v0 +at

x = x0 + v0t +1/2at²

Question 1

2.2 s after the start the hare is accelerating (0.8 m/s²).

from the equation:

v = v0 +at

replacing with the data:

v = 0 m/s + 0.8 m/s² * 2.2 s = 1.76 m/s

Question 2

At 15.1 s the hare is running at constant speed. It will be the final speed reached during the first 4.4 s:

v = v0 + a*4.4s

replacing with the data:

v= 0 m/s + 0.8 m/s² * 4.4s = 3.52 m/s

Question 3

We have to find the position at time 17.9 s.

For the first 4.4 s the hare runs:

x = x0 + v0t +1/2at² = 0m + 0 m/s * 4.4 s + 1/2 * 0.8 m/s² * (4.4 s)² = 7.7 m

For the next 13.5 s, the hare runs:

x = x0 + vt

where v=v0 + a*4.4s (the final velocity of the first 4.4 s)

v = 0 m/s + 0.8 m/s² * 4.4 s = 3.52 m/s

and x0 = 7.7 m (the final position of the first sprint)

Then:

x = 7.7m + 3.54 m/s * 13.5 s = 55.49 m

Question 4

The equation of position in this part of the race is:

x = x0 + v0t +1/2at²

where

x0 is the position calculated in question 3.

v0 is the final speed of the first 4.4 s calculated in question 2.

The velocity of the hare is 0 at position x = 61 m, then:

v = v0 +at

0 = v0 +at (at x = 61 m)

-v0 = at

a = -v0/t

then replacing a = -v0/t in the equation of position and solving for t:

x = x0 + v0t + 1/2(-v0/t)*t²

x = x0 + v0t -1/2v0t

x = x0 + 1/2v0t

x - x0 / (1/2v0) = t

replacing with the data:

61 m -55.49 m / 1/2* 3.53 m/s = 3.12 s

The acceleration is then:

a = -v0/t

a = -3.53 m/s / 3.12 s = -1.13 m/s²

Question 5

The hare moves for 4.4 s accelerating, for 13.5 s at a constant speed and for 3.12 s (see question 4) slowing down.

The total time is: 4.4s + 13.5 s + 3.12 s = 21.02 s

Question 6

The tortoise runs 61.0 meters in 21.02 s (the tortoise catches the hare just when it comes to stop). The equation for the position can be written as:

x = x0 +v0t +1/2at²

x0 = 0 and v0 = 0 since the tortoise starts from rest. Then, solving for a:

2x / t² = a

replacing with the data:

2*61 m / (21.02 s)² = a

a = 0.28 m/s²

"Slow and steady wins the race"