Cornell University

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Cornell University

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

If position is given by a function p(x), then the velocity is the first derivative of that function, and the acceleration is the second derivative. By using differential equations with either velocity or acceleration, it is possible to find position and velocity functions from a known acceleration.

I want to talk about position, velocity and acceleration and how differential equations can be used to show the relationships between these.

Imagine an object that’s moving along a straight line. If we know that its position at any time t is s of t, here’s it’s position s of t, it’s velocity v of t is the derivative of s of t and its acceleration is the derivative of velocity. So if you think about what happens as we go in this direction, we’re differentiating, right? Going downward we’re differentiating. The derivative of position is velocity, the derivative of velocity is acceleration.

So going in the reverse direction, and this is the way we’re often going to go to have to go in. Problems that we’re going to do in this topic. We’re going to have to anti-differentiate to go from acceleration to velocity, from velocity to position. So anti-differentiate. Let me show you what a sample problem might look like.

Consider this problem. An object is moving along a straight line with acceleration a of t equals 12t-6 feet per second squared. If v of 1 is 9 and s of 1 is 15 find v of t and s of t. Find the functions. So taking this strictly as a mathematical problem. Based on what I just said, acceleration is the derivative of velocity. I’d have to anti-differentiate to get velocity. Anyway, let’s set this up like a differential equation. If a of t equals 12t-6 and a of t is v prime of t, this is a differential equation, v prime of t equals 12t-6 subject to the initial conditions v of 1 equals 9 and s of 1 equals 15. So this is an initial value problem, you have a differential equation, 2 initial conditions and we’ll solve this in a future exercise.