AP Physics 1 Exam: Displacement

A car starts from rest and undergoes constant acceleration for 10 seconds. The car then maintains a constant velocity for 15 seconds, after which it decelerates uniformly and comes to a stop after another 5 seconds. The displacement of the car during this time period is recorded as 150 meters. Calculate the acceleration and maximum velocity of the car during the second phase (constant velocity).

Answer:

Let's analyze the motion of the car during the different phases:

Using the equation for displacement for uniformly accelerated motion:

Equation 1: Δx = v₀ * t + (1/2) * a * t^2

Where:

• Δx is the displacement
• v₀ is the initial velocity
• t is the time
• a is the acceleration

Phase I:

Since the car starts from rest, the initial velocity is zero. Using Equation 1, we have: Δx₁ = 0 * 10 + (1/2) * a * 10^2 = 50a Δx₁ = 50a

Phase II:

During this phase, the object maintains a constant velocity, meaning the acceleration is zero. The displacement is given as 150 meters, thus: Δx₂ = v₂ * t₂ = 150 v₂ * 15 = 150 v₂ = 10 m/s

Phase III:

During this phase, the car decelerates uniformly until it comes to a stop. The acceleration is -a, and the final velocity is zero. Using Equation 1, we have: Δx₃ = 0 * 5 + (1/2) * (-a) * 5^2 = -12.5a Δx₃ = -12.5a

Now, we need to relate the displacements:

Δx = Δx₁ + Δx₂ + Δx₃ 150 = 50a + 150 - 12.5a

Simplifying the equation: 150 = 37.5a + 150

Solving for acceleration: 37.5a = 0 a = 0 m/s^2

Therefore, the acceleration during the second phase (constant velocity) is zero.

To find the maximum velocity, we need to use the third phase. We know the final velocity is zero and the time is 5 seconds. Using Equation 1, we have: 0 = v₂ + (-a) * 5 v₂ = 5a v₂ = 5 * 0 v₂ = 0 m/s

Therefore, the maximum velocity during the second phase is 0 m/s.