AP Physics C Mechanics 1.2: Instantaneous Motion
Use calculus-based descriptions of position, velocity, and acceleration to analyze how an object’s motion changes at an instant.
- Syllabus
- Effective Fall 2025
- Course
- AP Physics C: Mechanics
Use calculus-based descriptions of position, velocity, and acceleration to analyze how an object’s motion changes at an instant.
Scientists have created a new type of lightweight foam and are performing experiments to investigate the properties of the foam. The mass of Cart A is 1000 kg and the mass of Cart B is 2000 kg. A piece of foam with negligible mass is attached to the front of Cart A, as shown. Cart A moves with a constant speed toward Cart B, which is initially at rest. At time t=0 s, the foam connected to Cart A makes contact with Cart B. The foam remains in contact with Cart B for 0.5 s, after which the carts separate and both carts move with constant velocities.

For 0≤t≤0.50 s, the velocity v of Cart A can be described by the function v(t)=64t3−48t2+5.
Calculate the magnitude of the maximum net force acting on Cart A during this interval.
On the following grid, draw a smooth curve of the magnitude of the force acting on Cart A as a function of time. Clearly indicate the value of the maximum force on the vertical axis.

The foam is removed from the front of Cart A and the experiment is repeated. The carts collide, with both Cart A and Cart B having the same initial and final velocities as in the original collision. The time intervals during which the carts are in contact are different in the collision with the foam and the collision without the foam. In the collision without the foam, Cart A is in contact with Cart B for a shorter duration than in the original collision, when the foam was present.
For the original collision when the foam is present, the magnitude of the average net force exerted on Cart B is F1. For the collision without the foam, the magnitude of the average net force exerted on Cart B is F2.
a(t)dtda=dtdv=192t2−96t,=384t−96=0⟹t=0.25 s.
At (t=0.25\ \mathrm{s}), (a=-12\ \mathrm{m\,s^{-2}}). With (m=1000\ \mathrm{kg}),
Fmax=m∣a∣=(1000)(12)=12000 N.
For the graph,
∣F(t)∣=1000∣192t2−96t∣=96000t(1−2t),0≤t≤0.50,
so it is a smooth downward-opening curve through (0\ \mathrm N) at (t=0) and (t=0.50\ \mathrm s), with maximum (12\,000\ \mathrm N) at (t=0.25\ \mathrm s).