AP Physics 1 Exam Question

A car is traveling on a straight road. Its initial velocity is 15 m/s and it comes to rest after traveling a distance of 100 meters. The car's acceleration is constant throughout its motion.

1. Calculate the car's acceleration.
2. Determine the time it takes for the car to come to rest.
3. Find the car's average velocity over the entire motion.
4. Now, suppose the car travels the same distance of 100 meters in twice the time. Calculate the acceleration required in this case.

Answer

1. Using the first equation of motion, we have:

   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 traveled.

   Since the car comes to rest, its final velocity (v) is 0 m/s. The initial velocity (u) is 15 m/s, and the distance traveled (s) is 100 meters. Substituting these values into the equation:

   (0)^2 = (15)^2 + 2a(100)

   Solving for acceleration (a), we get:

   0 = 225 + 200a -225 = 200a a = -225/200 a = -1.125 m/s^2

   Therefore, the car's acceleration is -1.125 m/s^2 (negative because the car is decelerating).

2. To determine the time it takes for the car to come to rest, we can use the second equation of motion:

   v = u + at

   Since the final velocity (v) is 0 m/s, the initial velocity (u) is 15 m/s, and the acceleration (a) is -1.125 m/s^2, substituting these values into the equation:

   0 = 15 + (-1.125)t

   Rearranging the equation and solving for time (t):

   1.125t = 15 t = 15 / 1.125 t ≈ 13.33 s

   Therefore, the car takes approximately 13.33 seconds to come to rest.

3. The average velocity (v_avg) over the entire motion can be calculated using the formula:

   v_avg = (u + v) / 2

   For this scenario, the initial velocity (u) is 15 m/s and the final velocity (v) is 0 m/s. Substituting these values into the equation:

   v_avg = (15 + 0) / 2 v_avg = 7.5 m/s

   Therefore, the car's average velocity over the entire motion is 7.5 m/s.

4. If the car travels the same distance of 100 meters in twice the time, we need to find the new acceleration required. Let's call it a'.

   Since the distance traveled remains the same, we can use the third equation of motion:

   s = ut + (1/2)at^2

   Let's denote the new time as 2t.

   100 = 15(2t) + (1/2)a'(2t)^2 100 = 30t + 2a' t^2

   Rearranging the equation:

   2a' t^2 + 30t - 100 = 0

   This is a quadratic equation. Solving for the unknown acceleration (a') using quadratic formula:

   t = (-30 ± √(30^2 - 4(2)(-100))) / (2(2)) t = (-30 ± √(900 + 800)) / 4 t = (-30 ± √(1700)) / 4

   Since we are considering positive time, we select the positive square root:

   t = (-30 + √(1700)) / 4 t ≈ 7.17 s

   Substituting this value back into the equation:

   100 = 30(7.17) + 2a'(7.17)^2 100 = 215.1 + 102.474a' -115.1 = 102.474a' a' ≈ -1.124 m/s^2

   Therefore, the required acceleration to travel the same distance in twice the time is approximately -1.124 m/s^2 (almost the same as the original acceleration).