Two cars start from rest at the same location
and at the same instant and race along a straight
track. Car A accelerates at 6.6 fps2 to a speed of
90 mph and then runs at this constant speed. Car B
accelerates at 4.4 fps2 to a speed of 96 mph and
then runs a a constant speed.
(a) Which car will with a 3-mile race and by what
distance?
(b) What will be the maximum lead of A over B?
(c) How far will the cars have traveled when B
passes A?
To determine the answers to these questions, we need to use the equations of motion and kinematics principles. Let's go through each question step by step:
(a) To determine which car will win the race and by what distance, we first need to find the time it takes for each car to reach their constant speeds.
For Car A:
Initial velocity (u) = 0 (since it starts from rest)
Acceleration (a) = 6.6 fps^2
Final velocity (v) = 90 mph
To convert mph to fps, we multiply by 1.47 since 1 mph = 1.47 fps.
Therefore, v = 90 * 1.47 = 132.3 fps.
Using the equation v = u + at and rearranging it to solve for time (t):
t = (v - u) / a
t = (132.3 - 0) / 6.6
t ≈ 20.05 seconds
So, it takes Car A approximately 20.05 seconds to reach its constant speed.
For Car B:
Initial velocity (u) = 0 (since it starts from rest)
Acceleration (a) = 4.4 fps^2
Final velocity (v) = 96 mph
To convert mph to fps, we multiply by 1.47 since 1 mph = 1.47 fps.
Therefore, v = 96 * 1.47 = 141.12 fps.
Using the same equation and substituting the values:
t = (v - u) / a
t = (141.12 - 0) / 4.4
t ≈ 32.07 seconds
So, it takes Car B approximately 32.07 seconds to reach its constant speed.
To find out which car wins the race, we compare the times taken for each car to reach their constant speeds. Since Car A takes less time (20.05 seconds) compared to Car B (32.07 seconds), Car A wins the race.
Now, let's find the distance by which Car A wins the race. We'll calculate the distance traveled by each car during this time.
For Car A:
Distance traveled (s) = ut + (1/2)at^2
s = 0*t + (1/2)*6.6*(20.05^2)
s ≈ 0 + 6.6*201 (1 fps^2 = 1 s^2)
s ≈ 1326 feet
For Car B:
Distance traveled (s) = ut + (1/2)at^2
s = 0*t + (1/2)*4.4*(32.07^2)
s ≈ 0 + 4.4*1030 (1 fps^2 = 1 s^2)
s ≈ 4522 feet (rounded to the nearest foot)
Therefore, Car A wins the race by approximately 4522 - 1326 = 3196 feet (rounded to the nearest foot).
(b) Now let's determine the maximum lead of Car A over Car B. To do this, we need to find out when Car B reaches its constant speed.
For Car B:
Initial velocity (u) = 0 (since it starts from rest)
Acceleration (a) = 4.4 fps^2
Final velocity (v) = 96 mph
Using the same conversion factor, v = 96 * 1.47 = 141.12 fps.
Using the same equation as before, we can calculate the time taken by Car B to reach its constant speed:
t = (v - u) / a
t = (141.12 - 0) / 4.4
t ≈ 32.07 seconds
Hence, it takes Car B approximately 32.07 seconds to reach its constant speed.
Since Car A will be running at its constant speed of 90 mph (132.3 fps) during this time, the maximum lead of Car A over Car B will be the distance traveled by Car A during the time it takes for Car B to reach its constant speed.
To find the distance traveled by Car A during this time:
Distance traveled (s) = ut
s = 132.3 * 32.07
s ≈ 4247 feet (rounded to the nearest foot)
Therefore, the maximum lead of Car A over Car B is approximately 4247 feet (rounded to the nearest foot).
(c) Lastly, let's determine how far the cars will have traveled when Car B passes Car A. At this point, both cars will be running at their constant speeds, so we can set up an equation to find the time it takes for Car B to catch up with Car A.
The relative speed of Car B with respect to Car A is the difference in their constant speeds:
Relative speed = velocity of Car B - velocity of Car A
Relative speed = 141.12 - 132.3
Relative speed ≈ 8.82 fps
Now, we can set up the equation: Distance = relative speed * time
Let's assume the distance traveled when Car B passes Car A is d. The time taken for this to happen is the same for both cars.
For Car A:
Distance traveled (s) = velocity of Car A * time
s = 132.3 * t
For Car B:
Distance traveled (s) = velocity of Car B * time
s = 141.12 * t
Now we equate these two distances, as both cars meet at the same distance:
132.3 * t = 141.12 * t
By canceling out t from both sides, we find that Car B will pass Car A at the same distance traveled. Therefore, they will have both traveled the same distance.
Hence, when Car B passes Car A, both cars will have traveled the same distance.