A motorist traveling with constant speed of vc = 16.0 m/s passes a school-crossing corner, where the speed limit is 10 m/s. Just as the motorist passes, a police officer on a motorcycle stopped at the corner starts off in pursuit. The officer accelerates from rest at am = 2.67 m/s2 until reaching a speed of 26.7 m/s. The officer then slows down at a constant rate until coming alongside the car at x = 240.0 m.
Consider a coordinate system with origin at the school-crossing corner, x=0, and the +x-axis in the direction of the car's motion.
(a) How long does it take for the motorcycle to catch up with the car (in s)?
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(b) How long does it take for the motorcycle to speed up to 26.7 m/s? (Express your answer in s.)
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(c) How far (in m) is the motorcycle from the corner when switching from speeding up to slowing down?
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(d) How far (in m) is the motorcycle from the car when switching from speeding up to slowing down?
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(e) What is the acceleration (in m/s2) of the motorcycle when slowing down? (pay attention to the sign)
To solve this problem, we need to analyze the motion of both the car and the motorcycle and determine the time and distance at different points.
First, let's look at part (a): How long does it take for the motorcycle to catch up with the car?
We can use the following equation to calculate the time it takes for the motorcycle to catch up with the car:
d = ut + (1/2)at^2
Where d is the distance, u is the initial velocity, a is the acceleration, and t is the time.
For the car, its initial velocity is 16.0 m/s, and it travels a distance of x = 240.0 m.
d_car = 16.0t_car (since the car travels at a constant speed)
For the motorcycle, it starts from rest and accelerates at 2.67 m/s^2 until it reaches a speed of 26.7 m/s.
We can calculate the time it takes for the motorcycle to reach this speed using the equation:
v = u + at
26.7 = 0 + 2.67t_motorcycle
Solving for t_motorcycle, we find:
t_motorcycle = 26.7 / 2.67 = 10 s
Now, we can calculate the total time it takes for the motorcycle to catch up with the car. Since the motorcycle and the car are moving at the same time and the same distance, we can set up the equation:
t_car = t_motorcycle
16.0t_car = 10
Solving for t_car, we find:
t_car = 10 / 16.0 = 0.625 s
Therefore, it takes the motorcycle 0.625 seconds to catch up with the car.
Moving on to part (b): How long does it take for the motorcycle to speed up to 26.7 m/s?
We have already found the time it takes for the motorcycle to reach a speed of 26.7 m/s, which is 10 seconds.
Now, we can move on to parts (c) and (d):
(c) How far is the motorcycle from the corner when switching from speeding up to slowing down?
Since the motorcycle is accelerating at a constant rate until it reaches a speed of 26.7 m/s, we can use the equation:
v^2 = u^2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Plugging in the values, we have:
26.7^2 = 0 + 2 * 2.67 * s_motorcycle
Solving for s_motorcycle, we find:
s_motorcycle = (26.7^2) / (2 * 2.67) = 267 m
Therefore, the motorcycle is 267 meters from the corner when switching from speeding up to slowing down.
(d) How far is the motorcycle from the car when switching from speeding up to slowing down?
Since the car and the motorcycle are moving at the same time and the same distance, we can simply use the distance of the car:
s_car = 240 m
Therefore, the motorcycle is 240 meters from the car when switching from speeding up to slowing down.
Finally, for part (e): What is the acceleration of the motorcycle when slowing down?
We can calculate the acceleration when slowing down using the equation:
a = (v - u) / t
Where v is the final velocity, u is the initial velocity, and t is the time.
From the information given, the motorcycle slows down from 26.7 m/s to a final velocity of 16.0 m/s. The time it takes for this deceleration is the same as the time it takes for the motorcycle to catch up with the car, which we found to be 0.625 seconds.
Plugging in the values, we have:
a = (16.0 - 26.7) / 0.625 = -16.8 m/s^2
Therefore, the acceleration of the motorcycle when slowing down is -16.8 m/s^2, with the negative sign indicating deceleration.
To solve this problem, we can use the equations of motion to find the required values. Let's calculate the answers step by step:
(a) How long does it take for the motorcycle to catch up with the car?
We need to find the time taken by the motorcycle to reach the position of the car.
First, let's find the time taken by the motorcycle to accelerate to its final velocity of 26.7 m/s using the equation:
v = u + at
Here, v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.
Given:
Final velocity, vf = 26.7 m/s
Acceleration, a = 2.67 m/s^2
Using the equation, we can rearrange it to solve for t:
t = (vf - u) / a
t = (26.7 m/s - 0) / 2.67 m/s^2
t ≈ 10 s*
So, it takes approximately 10 seconds for the motorcycle to speed up to 26.7 m/s.
(b) How long does it take for the motorcycle to speed up to 26.7 m/s?
We already calculated this in the previous step. It takes approximately 10 seconds for the motorcycle to speed up to 26.7 m/s.
(c) How far is the motorcycle from the corner when switching from speeding up to slowing down?
To find this distance, we need to use the kinematic equation:
s = ut + 0.5at^2
Here, s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity, ui = 0 m/s (since the motorcycle starts from rest, it has an initial velocity of 0)
Acceleration, a = 2.67 m/s^2
Time, t = 10 seconds*
Plugging in the values:
s = 0 * 10 + 0.5 * 2.67 * (10)^2
s ≈ 133.5 m*
So, the motorcycle is approximately 133.5 meters from the corner when switching from speeding up to slowing down.
(d) How far is the motorcycle from the car when switching from speeding up to slowing down?
We are given that the motorist is at x = 240.0 m from the corner. So, the distance between the motorcycle and the car when the motorcycle switches from speeding up to slowing down is:
Distance = Car's position - Motorcycle's position
Distance = 240.0 m - 133.5 m
Distance ≈ 106.5 m*
So, the motorcycle is approximately 106.5 meters away from the car when switching from speeding up to slowing down.
(e) What is the acceleration of the motorcycle when slowing down?
To find the acceleration when slowing down, we can use the same equation as in part (c):
s = ut + 0.5at^2
We are given:
Final velocity, vf = 26.7 m/s
Initial velocity, ui = 0 m/s
Time, t = 10 seconds*
Plugging in the values:
s = 0 * 10 + 0.5 * a * (10)^2
s = 0 + 5a
Since the motorist and the motorcycle reach the same position, s = 240.0 m.
240 = 5a
Solving for a:
a = 240 / 5
a = 48 m/s^2*
So, the acceleration of the motorcycle when slowing down is 48 m/s^2 (pay attention to the negative sign, which indicates deceleration since the motorcycle is slowing down).
*Note: The values provided are approximations, rounded to the nearest decimal point.