As high school students move into their physics courses, especially in AP Physics 1, they’re going to encounter problems involving the study of kinematics, or one-dimensional motion. One of the typical kinds of problems that an instructor may ask students to solve involves acceleration and the application of Newton’s Laws of Motion. Here’s an example.

A 43.0 kg parachutist is moving straight downward with a speed of 4.00 meters per second. If the parachutist comes to rest after constantly accelerating over a distance of meters, what force does the ground exert on her?

We are ultimately trying to calculate the force of the ground on the parachutist, which means we’re going to use Newton’s Second Law of Motion. To be able to use that equation, we need to figure out the acceleration of the parachutist.

To determine the acceleration, we will use this formula:

v_f² – v_i² = 2ax

where

v_f = final velocity = 0 (the parachutist stops when he hits the ground)

v_i = initial velocity = 4.00 m/s (given)

a = acceleration

x = displacement = 0.785 m (given)

Since we are solving for acceleration, we’ll move the variables around before we plug in what we know.

a = v_f²– v_i² / 2x

a = 0 – 16 / 2 (0.785) = – 16 / 1.57 = 10.19 m/s²

Well, okay, but the problem is that we know the acceleration of gravity on Earth is 9.81 m/s². So the acceleration can’t be faster than that.

Now we can figure out the force acting on the parachutist.

F = ma

F = (43.0 kg) (9.81 m/s²) = – 421.83 N

Here at Peak Tutoring, we can help your child master physics and, if she is on the adventure of AP Physics, help her excel on the exam.