Position, Velocity and Acceleration - Problem 2

Explanation

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.

Transcript

Let’s do a slightly harder problem that involves position velocity and acceleration. A car manufacture claims its model accelerates form 0 to 120 miles per hour in 1 minute. If its acceleration is constant, how far will the car travel by the time it reaches 120 miles per hour?

Before I figure out how to solve this problem, let me just write down the information that I’m given.

The fact that the acceleration is constant is really important. So I could write a(t) is equal to k. And acceleration is also the derivative of velocity. So I could wire that as v'(t) equals k. This is a differential equation. What about the other pieces of information? Initially, its velocity is 0, its final velocity is 120 miles per hour, and it does this is in 1 minute.

Notice the units of miles in hours through. This is going to be really critical. If my units are hours, for time, and miles for distance, so I have to stick with that throughout the problem or else change over. I’m going to choose to stick with them and that means that time, t, is in hours. So my initial conditions are v(0) equals 0. So when t equals 0, the velocity is 0. But my second condition is that after 1 minute, I’m at 120. I minute is 1/60 of an hour. These are two initial conditions.

I have no information about position but I could make the position function; s(t), be the position relative to the starting point, meaning that when t equals 0, s is 0. It doesn’t really matter but I want to know how far the car has travelled. So if I make s relative to the starting point, then s(1/60) would be what I’d be looking for in the final answer. So let’s make s(0) equal to 0. This will be another initial condition.

So we’re focusing on this equation. That’s my differential equation. And I solve it by anti differentiating. V(t) equals the integral of kdt. Remember my independent variable’s time here so I have to integrate with respect to t. I get k times t plus a constant. So two parameters here, I don’t know what k is and I don’t know what c is but I can solve for at least some of that information by using an initial condition. Let me use v(0) equals 0.

V(0) equals 0 means, when I plug 0 in for t, I get 0 velocity. So 0 for this equal k times 0 plus c. this tells me that c is 0. That tells me that my velocity function is actually k times t.That's good to know. Now let’s take this and write it as a differential equation because v(t) is the same as s'(t). So s'(t) equals k times t. And if I want to find s(t) I integrate. S(t) is the integral of k times t dt.

Now the integral of t is 1/2t² so k times t is going to be k over 2, t² plus a constant. I’ll use a different letter this time, D. So that’s my position function. Remember that s(0) equals 0. My position is relative to where I started when I first started accelerating. S(0) equals 0 means I plug in 0 for position when time is 0. So 0 for position equals, k over 2 times 0² plus D, that tells me that D equals 0. D equals 0. My position function is k over 2t². I still need to figure out what k is. What that constant of acceleration is.

There’s a piece of information that I haven’t used yet, and that’s that v(1/60) equals 120. Remember after a minute I went 120 miles per hour. I could have used that here. V(t) equals k times t. So let me use it now. V of 1/60 equals 120. So this is 120, when t is 1/60. 120 equals k times 1/60. So to find k, I just multiply both sides of this equation by 60. I get 7200 equals k. So this value, plug it back in here, k over 2, I get 3600t². This is my position function.

Now let’s go back to the original problem just really quickly. If acceleration is constant, how far will the car travel by the time it reaches 120 miles per hour? We know that that time is going to be 1/60. So if I plug 1/60 into this position, I should get how far it travelled. Remember, my starting point was s equals 0. S(1/60) is 3600 times (1/60)². (1/60)² is 1/3600. So these will cancel each other and leave 1.

What are the units? My units for velocity were miles per hour and when I integrate miles per hour with respect to hours, I get position in miles. This will be in miles. We’ve gone 1 mile. I accelerated from 0 to 120 in 1 minute and we’ll have travelled 1 mile while doing that.

Tags
differential equations derivatives position velocity acceleration