 ###### Norm Prokup

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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# Position, Velocity and Acceleration - Problem 1

Norm Prokup ###### Norm Prokup

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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Position, velocity, and acceleration problems can be solved by solving differential equations. Acceleration is the derivative of velocity, and velocity is the derivative of position. So, to find the position function of an object given the acceleration function, you'll need to solve two differential equations and be given two initial conditions, velocity and position.

Since a(t)=v'(t), find v(t) by integrating a(t) with respect to t. Solve for c by using the initial velocity condition. Then, since v(t)=x'(t), solve for the position function x(t) by integrating the answer you found for v(t) with respect to t. Again, solve for c by using the initial position condition. The result is the position function. Given the position function, you can solve for the position of an object given the time t.

Let’s do that problem we just described before. An object is moving along a straight line. Its acceleration is given by a(t) equals 12t minus 6. V(1) is 9, s(1) is 15. We need to find the velocity and position functions.

Now remember, when you’re dealing with position, velocity and acceleration, these problems are often just differential equations problems. All you need to know is the relationship between position, velocity and acceleration. Take a look up here. I’ve stacked them up, top to bottom.

Positions is s(t). Velocity is the derivative of s and acceleration is the derivative of velocity. So going downward, we differentiate to get from position to velocity, from velocity to acceleration. Going backwards, which is what we’re going to have to do, we antidifferentiate. We antidifferentiate acceleration to get velocity. We antidifferentiate velocity to get position. So remember this set up and you’ll always remember whether you need to differentiate or antidifferentiate to get from one to the other.

We have acceleration. And we know that acceleration’s the same as the derivative of velocity. so I can write v'(t) equals 12t minus 6. That’s just a differential equation. I solve it by integrating. So velocity is the integral of 12t minus 6. Now very important, it’s the integral with respect to t. Here my independent variable is t, so all derivatives are going to be derivatives with respect to t, and all integrals will be integrals with respect to t.

When I integrate this I get v(t) equals the integral of 12t is 12 times the integral of t, which is 1/2t². So this will be 6t² and the integral of 6 is just 6t. And of course, I have to throw in a constant. Now I can find the value of that constant using the initial condition v(1) equals 9. So let me use that now.

If v(1) equals 9, then that means that when I plug 9 in for v(t), I should plug 1 in for t. So 9 equals 6 times 1² minus 6 times 1 plus 6. These 6s are just going to cancel and I get that c is 9. So my velocity function is v(t) equals 6t² from here minus 6t plus c which we now know is 9. That’s my velocity function. So I’m half done.

Let me write this up here. Velocity equals 6t² minus 6t plus 9. This also can be written as a differential equation because velocity is the derivative of position. S'(t) equals 6t² minus 6t plus 9. And we solve this by antidifferentiating so s(t) will equal the integral of this whole thing with respect to t.

6t² minus 6t plus 9dt. That means that s(t) equals; now what’s the antiderivative of 6t². The antiderivative of t² is 1/3t³. So 6 times that is 2t³. What’s the antiderivative of minus 6t? The antiderivative of t is 1/2t² times -6 is -3t² and the antiderivative of 9 is just 9t. And I should add a constant here, but I’m going to add a different constant than I did last time. I used c last time, so I’ll use D this time. And this is my position function.

To find D I need to use the initial condition which was s(1) equals 15. When I plug ion 1 for t, I need to plug in 15 for the position. So 15 equals 2 times 1³, minus 3 times 1² plus 9 times 1 plus D. 2 minus 3 plus 9 plus D. So 15 is 2 minus 3 plus 9 plus D. I look at this I get -1, plus 9 is 8, so this is just 8. And so D has to be 7. My final answer for position is s(t) equals 2t³ minus 3t² plus 9t plus 7. That’s my position, this is my velocity.

Just remember, the relationship between position and velocity and acceleration; in the down direction I’m differentiating, in the upward direction I’m antidifferentiating. If you’re starting out with acceleration and velocity, and asked for position, you’ve got to antidifferentiate.