A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 16.0m/s when it reaches the end of the ramp, which has length 118m.
A. What is the acceleration of the car?
B. How much time does it take the car to travel the length of the ramp?
C. The traffic on the freeway is moving at a constant speed of 16.0 . What distance does the traffic travel while the car is moving the length of the ramp?
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I got 14.75 to be the time it took him to travel the ramp but that isn't the correct answer.
I'm not sure how to set up for the other parts.
To solve this problem, we can use the kinematic equations of motion. The key to solving the problem is understanding that the initial velocity of the car is zero (since it starts from rest) and the final velocity is 16.0 m/s.
Let's break down each part of the question:
A. What is the acceleration of the car?
To find the acceleration, we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (16.0 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
s = displacement (118 m)
Rearranging the equation to isolate the acceleration, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values, we get:
a = (16.0^2 - 0^2) / (2 * 118)
Calculating the expression gives us:
a ≈ 0.116 m/s^2
So, the acceleration of the car is approximately 0.116 m/s^2.
B. How much time does it take the car to travel the length of the ramp?
To find the time, we can use the equation:
v = u + at
where:
v = final velocity (16.0 m/s)
u = initial velocity (0 m/s)
a = acceleration (0.116 m/s^2, as found in part A)
t = time (unknown)
Rearranging the equation to isolate the time, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (16.0 - 0) / 0.116
Calculating the expression gives us:
t ≈ 138 seconds
So, it takes approximately 138 seconds for the car to travel the length of the ramp.
C. The traffic on the freeway is moving at a constant speed of 16.0 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
To find this distance, we need to calculate how much distance the car travels on the ramp. We can use the equation:
s = u * t + (1/2) * a * t^2
where:
s = distance (unknown)
u = initial velocity (0 m/s)
t = time (138 seconds, as found in part B)
a = acceleration (0.116 m/s^2, as found in part A)
Plugging in the values, we get:
s = 0 * 138 + (1/2) * 0.116 * (138)^2
Calculating the expression gives us:
s ≈ 1,121.1 meters
So, the car travels approximately 1,121.1 meters while on the ramp.
To find the distance that the traffic travels while the car is on the ramp, we need to calculate the difference between the total distance of the ramp and the distance traveled by the car.
Total distance of the ramp = 118 meters (given in the problem)
Distance traveled by the car = 1,121.1 meters (as found above)
Distance traveled by the traffic = Total distance of the ramp - Distance traveled by the car
= 118 - 1,121.1
≈ -1,003.1 meters
The negative value indicates that the traffic on the freeway is moving backward relative to the car on the ramp. Therefore, the traffic does not travel any positive distance while the car is on the ramp.