A motorist traveling with constant speed of vc = 17.000 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 = 3.190 m/s2 until reaching a speed of 25.500 m/s. The officer then slows down at a constant rate until coming alongside the car at x = 346.000 m, traveling with the same speed as the car.
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)?
(b) How long does it take for the motorcycle to speed up to 25.500 m/s? (Express your answer in s.)
(c) How far (in m) is the motorcycle from the corner when switching from speeding up to slowing down?
(d) How far (in m) is the motorcycle from the car when switching from speeding up to slowing down?
(e) What is the acceleration (in m/s2) of the motorcycle when slowing down? (pay attention to the sign)
To answer these questions, we need to analyze the motion of both the car and the motorcycle. Let's break it down step by step.
Step 1: Determine the time it takes for the motorcycle to catch up with the car.
To find the time it takes for the motorcycle to catch up with the car, we need to analyze their positions as functions of time.
For the car:
The car is traveling at a constant speed of vc = 17.000 m/s. The distance it travels from the school-crossing corner (x = 0) can be calculated using the formula:
x_car = vc * t,
where x_car is the position of the car and t is the time.
For the motorcycle:
The motorcycle starts from rest and accelerates at am = 3.190 m/s^2 until reaching a speed of 25.500 m/s. The distance it travels can be calculated using the formula for uniformly accelerated motion:
x_motorcycle = 0.5 * am * t^2,
where x_motorcycle is the position of the motorcycle and t is the time.
To find the time at which the motorcycle catches up with the car, we equate the distances traveled by both:
vc * t = 0.5 * am * t^2.
Solving for t, we get:
0.5 * am * t^2 - vc * t = 0.
This is a quadratic equation in t. Solving it will give us the time it takes for the motorcycle to catch up with the car.
Step 2: Determine the time it takes for the motorcycle to speed up to 25.500 m/s.
To find the time it takes for the motorcycle to speed up, we use the formula for uniformly accelerated motion:
v = u + a * t,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the motorcycle starts from rest, so u = 0 m/s. We are given that the final velocity is v = 25.500 m/s, and the acceleration is am = 3.190 m/s^2. Solving for t, we can find the time it takes for the motorcycle to speed up.
Step 3: Determine the distance at which the motorcycle switches from speeding up to slowing down.
The distance at which the motorcycle switches from speeding up to slowing down is given by the equation:
x_motorcycle = 0.5 * am * t^2,
where x_motorcycle is the position of the motorcycle and t is the time it takes for the motorcycle to speed up.
Step 4: Determine the distance between the motorcycle and the car when switching from speeding up to slowing down.
To find the distance between the motorcycle and the car when switching from speeding up to slowing down, we calculate the difference in their positions at that time:
x_diff = x_car - x_motorcycle.
Step 5: Determine the acceleration of the motorcycle when slowing down.
The acceleration of the motorcycle when slowing down is given by the formula:
a_motorcycle = (v_motorcycle - u_motorcycle) / t,
where a_motorcycle is the acceleration, v_motorcycle is the final velocity, u_motorcycle is the initial velocity, and t is the time it takes for the motorcycle to speed up.
Now, we can calculate the answers to the specific questions:
(a) To find the time it takes for the motorcycle to catch up with the car, solve the quadratic equation 0.5 * am * t^2 - vc * t = 0 for t.
(b) To find the time it takes for the motorcycle to speed up to 25.500 m/s, use the formula v = u + a * t with v = 25.500 m/s and a = am.
(c) To find the distance at which the motorcycle switches from speeding up to slowing down, use the equation x_motorcycle = 0.5 * am * t^2 with t from part (b).
(d) To find the distance between the motorcycle and the car when switching from speeding up to slowing down, calculate the difference x_diff = x_car - x_motorcycle.
(e) To find the acceleration of the motorcycle when slowing down, use the formula a_motorcycle = (v_motorcycle - u_motorcycle) / t with t from part (b).
To solve this problem, we can use kinematic equations to calculate the time, distance, and acceleration involved in the scenario. Let's break down the problem and solve it step-by-step.
Given:
vc = 17.000 m/s (speed of the car)
vlimit = 10 m/s (speed limit)
am = 3.190 m/s^2 (acceleration of the motorcycle)
vm = 25.500 m/s (final speed of the motorcycle)
x = 346.000 m (distance when the motorcycle comes alongside the car)
We'll use the following kinematic equation to solve the problem:
v = u + at
s = ut + (1/2)at^2
Step 1: Find the time for the motorcycle to catch up with the car.
The motorcycle catches up with the car when their speeds are equal.
Therefore, we can equate the speed of the car and motorcycle as follows:
vc = vm
Substituting the given values:
17.000 m/s = vm
Therefore, it takes 17 seconds for the motorcycle to catch up with the car.
Answer:
(a) It takes 17 seconds for the motorcycle to catch up with the car.
Step 2: Find the time for the motorcycle to speed up to 25.500 m/s.
We'll use the following kinematic equation to solve for time:
v = u + at
Substituting the given values:
vm = am * t
25.500 m/s = 3.190 m/s^2 * t
Solving for t:
t = 25.500 m/s / 3.190 m/s^2
Therefore, it takes approximately 8 seconds for the motorcycle to speed up to 25.500 m/s.
Answer:
(b) It takes approximately 8 seconds for the motorcycle to speed up to 25.500 m/s.
Step 3: Find the distance from the corner when the motorcycle switches from speeding up to slowing down.
We'll use the following kinematic equation to solve for distance:
s = ut + (1/2)at^2
Substituting the given values:
s = 0 * t + (1/2)am * t^2
Simplifying:
s = (1/2)am * t^2
Substituting the values:
s = (1/2) * 3.190 m/s^2 * (8 s)^2
Therefore, the distance from the corner when switching from speeding up to slowing down is approximately 102.08 meters.
Answer:
(c) The distance from the corner when switching from speeding up to slowing down is approximately 102.08 meters.
Step 4: Find the distance between the motorcycle and the car when the motorcycle switches from speeding up to slowing down.
We know the distance traveled by the car when the motorcycle catches up with it is 346.000 m (given value).
Therefore, the distance between the motorcycle and the car when switching from speeding up to slowing down is:
d = 346.000 m - distance from the corner to the switching point
Substituting the values:
d = 346.000 m - 102.08 m
Therefore, the distance between the motorcycle and the car when switching from speeding up to slowing down is approximately 243.92 meters.
Answer:
(d) The distance between the motorcycle and the car when switching from speeding up to slowing down is approximately 243.92 meters.
Step 5: Find the acceleration (m/s^2) of the motorcycle when slowing down.
The acceleration when slowing down is negative because it is opposing the motion.
The final speed of the motorcycle is vm = 25.500 m/s (given value).
The initial speed of the motorcycle is u = 0 (since it starts from rest).
We'll use the following kinematic equation to solve for acceleration:
v^2 = u^2 + 2as
Substituting the given values:
(0 m/s)^2 = (25.500 m/s)^2 + 2a * 102.08 m
Solving for a:
a = -((25.500 m/s)^2) / (2 * 102.08 m)
Therefore, the acceleration of the motorcycle when slowing down is approximately -3.190 m/s^2.
Answer:
(e) The acceleration of the motorcycle when slowing down is approximately -3.190 m/s^2.