No type of calculus problem uses more concepts that are learned in the course more than that of uniform motion. For certain problems, a student may need to find the derivative. For other problems, students may need to integrate and solve at a point in time using an initial condition. Another benefit of learning how to solve uniform motion problems is that these problems have daily use outside of the calculus world (ex. speed of a pitch in baseball, acceleration from 0-60 in car tests, distance from point A to point B).

Let's get an example...

Example 1: A ball is thrown downward from a height of 512 feet with a velocity of 64 feet per second. How long will it take for the ball to reach the ground?
From the given conditions, you find that

So the ball will reach the ground 4 seconds after it is thrown.

Graphs of Uniform Motion:

The graph above is the position equation which equals s(t)= -16t^2+16t+32

The graph above is the velocity equation which is v(t)= -32t+16

The graph above is the Acceleration graph which is
a(t)= -32

Average velocity is the slope of secant through the position equation

## Uniform Motion

## By: James Colburn, Shane Nace, Cavan O'Reilly

No type of calculus problem uses more concepts that are learned in the course more than that of uniform motion. For certain problems, a student may need to find the derivative. For other problems, students may need to integrate and solve at a point in time using an initial condition. Another benefit of learning how to solve uniform motion problems is that these problems have daily use outside of the calculus world (ex. speed of a pitch in baseball, acceleration from 0-60 in car tests, distance from point A to point B).

Let's get an example...

Example 1:A ball is thrown downward from a height of 512 feet with a velocity of 64 feet per second. How long will it take for the ball to reach the ground?From the given conditions, you find that

So the ball will reach the ground 4 seconds after it is thrown.

Graphs of Uniform Motion:

The graph above is the position equation which equals s(t)= -16t^2+16t+32

The graph above is the velocity equation which is v(t)= -32t+16

The graph above is the Acceleration graph which is

a(t)= -32

Average velocity is the slope of secant through the position equation

Speed =lv(t)l

Example 2

Example 3

Links to Other Practice Problems