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"
