The brakes on a parked car fall and it accelerates at a constant rate, rolling into a lamp post down the hill. the car reaches a velocity of 4.3m/s by the time it smashes into the lamp post. The time between the brakes failing and the smashing of the car is 3.7s.
A. what is the cars acceleration?
b. What was the average velocity of the car while it was rolling?
C. how far did the car roll before the collision?
Give me answer
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To find the answers to the questions, we can solve the problem using the physics equations of motion. The relevant equation in this case is:
vf = vi + at
where:
vf is the final velocity,
vi is the initial velocity,
a is the acceleration, and
t is the time taken.
Let's address each question one by one:
A. What is the car's acceleration?
To find the car's acceleration, we can rearrange the equation to solve for acceleration:
a = (vf - vi) / t
Given that the initial velocity, vi, is 0 (since the car was parked), the final velocity, vf, is 4.3 m/s, and the time, t, is 3.7 s:
a = (4.3 m/s - 0) / 3.7 s
a = 1.16 m/s^2
Therefore, the car's acceleration is 1.16 m/s^2.
B. What was the average velocity of the car while it was rolling?
To find the average velocity, we can use the formula:
average velocity = (vi + vf) / 2
Given that the initial velocity, vi, is 0 (since the car was parked) and the final velocity, vf, is 4.3 m/s:
average velocity = (0 + 4.3 m/s) / 2
average velocity = 2.15 m/s
Therefore, the average velocity of the car while it was rolling is 2.15 m/s.
C. How far did the car roll before the collision?
To find the distance the car rolled, we can use the formula:
distance = vi * t + 0.5 * a * t^2
Given that the initial velocity, vi, is 0 (since the car was parked), the time, t, is 3.7 s, and the acceleration, a, is 1.16 m/s^2:
distance = 0 * 3.7 s + 0.5 * 1.16 m/s^2 * (3.7 s)^2
distance = 10.21 m
Therefore, the car rolled a distance of 10.21 meters before the collision.