A car moves at 12 m/s at the bottom of a hill and coasts up the hill with a uniform acceleration of -2.0 m/s. (A) Determine the time that the car stops and turns around and rolls back down the hill to its original position. (B) Determine the displacement during the first 1 second of coasting up the hill.
To solve these problems, we'll need to use the equations of motion.
(A) Determine the time that the car stops and turns around and rolls back down the hill to its original position:
1. We are given the initial velocity of the car at the bottom of the hill, v₀ = 12 m/s.
2. The car is coasting up the hill with a uniform acceleration, a = -2.0 m/s². Since the acceleration is negative, it means the car is slowing down.
3. We need to find the time it takes for the car to come to a stop. At that point, the velocity will be 0 m/s, and then it will start moving in the opposite direction.
4. We can use the equation v = v₀ + at to solve for the time it takes to stop the car.
0 = 12 + (-2)t (since v = 0)
5. Solve the equation for t:
-12 = -2t
Divide both sides by -2:
t = 6 seconds.
So, it takes 6 seconds for the car to stop and turn around.
(B) Determine the displacement during the first 1 second of coasting up the hill:
1. We need to determine the displacement of the car during the first 1 second while coasting up the hill.
2. To do this, we can use the equation s = s₀ + v₀t + (1/2)at², where s is the displacement, s₀ is the initial position (which we can assume to be 0), v₀ is the initial velocity, t is the time, and a is the acceleration.
3. Plug in the values we know: s₀ = 0, v₀ = 12 m/s, t = 1 second, and a = -2.0 m/s².
s = 0 + 12(1) + (1/2)(-2)(1)²
4. Solve the equation for s:
s = 12 + (-1) (simplifying)
s = 11 meters.
So, during the first 1 second of coasting up the hill, the car has a displacement of 11 meters.