Position, Velocity and Acceleration - Problem 3

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.

One special case is that of free fall motion. The acceleration for an object in free fall is a constant, -32 ft/s2 or -9.8 m/s2. With this information, you can solve for the velocity and position functions with the same techniques as described.

Transcript

We’re talking about position velocity and acceleration. And no discussion ion this topic is ever complete if we don’t talk about free fall. An object in free fall accelerates downward at a constant rate. This rate is approximately 32feet per second squared. This is the fact that we’re going to use in the next problem. It makes some assumptions; it assumes that you’re not falling from too high like from out of space. And of course it assumes that you’re falling on the earth. But yes this constant, 32 feet per second squared is very important.

Here is the problem. A ball is thrown upwards from 6 feet above ground with an initial velocity of 95 feet per second. Find a, how long before the ball hits the ground and b, the speed at which it strikes the ground.

First of all let’s observe that this is a free fall situation, but it’s a free fall situation with some initial conditions. First, the ball’s thrown upwards from 6 feet above the ground with an initial velocity of 95 feet per second. It’s also important that it’s thrown upwards. We’re going to need to distinguish between the upward direction and downward direction in this problem.

Let me draw a little diagram of what’s happening here. So imagine that this is the ground, and say this is a distance of 6 feet. So the ball is launched from this height. It goes up and then comes back down. Of course it probably comes back down right over the path it goes up if it’s thrown straight up. But for the purposes of illustration I’ll draw it this way. This is 6. Up ward direction, downward direction.

Now it’s very important that the initial velocity is up. If the initial velocity were down, it’s going to strike the ground very soon. So we need to make the distinction between those two cases. How do we do that? We have to pick a direction and assign it as the positive direction. In this problem I’m going to assign the upward direction a positive. That means the downward direction’s negative.

The reason this is important is, that acceleration is going to be in the downward direction. So the acceleration of 32 feet per second squared is going to be -32 feet per second squared. Very important that the acceleration is in the opposite direction of the initial velocity and the initial position. These are both positive. So let’s start the problem.

In free fall you have this acceleration of -32 feet per second squared, a downward acceleration. And acceleration is the same as the derivative of velocity. So we have a differential equation and we’ll solve this equation in a moment. Now what else do we have?

We have v(0) is 95. And we have that s(0) is 6. We have to initial conditions to go with our differential equation. So as we solve we’ll apply these initial conditions and find the values of constants. First we solve this differential equation by integrating. V(t) is the integral of -32dt. And the integral of a constant is a constant times t, and so we’ll get -32t plus c.

Now what’s the c value? Well I use the initial condition v(0) equals 95. I know that the velocity is 95 when t is zero. So when t is zero I get -32 times zero plus c. That means that this c value is 95. So now I know that my velocity is -32t plus 95. Let me write that up here. Now this is the same as s'(t) equals -32t plus 95. I solve this by integrating it.

S(t) equals the integral of -32t plus 95. What’s the anti derivative of -32t? The anti derivative of t is 1/2t², so times -32 gives me -16t². The anti derivative of the constant 95 is 95t, so plus 95t. And I need to add a new constant, D for my integration. This is my position function.

Now I need to figure out what D is. Remember my initial position s(0) equals 6. The ball was thrown from 6 feet above ground so when t equals zero, position is 6. So I’ll have 6 equals -16(0)² plus 95 times zero plus d. that tells me these are just zero. It tells me that D is 6. And I have the position function s(t) equals -16t² plus 95t plus 6. Now have I answered any of my questions yet?

Let’s take a look. It says how long before the ball hits the ground. Haven’t got that yet. B, the speed at which it strikes the ground, I haven’t found that yet either. I need to find a before b because I need to know when it strikes the ground in order to get the speed with which it strikes the ground.

How long before it strikes the ground? I can now answer that. Since I know the position, I can figure out when the position is zero, because that’s when it will hit the ground. This is the height above ground. So hits ground when s(t) is zero. Let me set this equal to zero and solve.

I get -16(t), this can actually be factored. This equation t. To get a 6 I can go with -1, -6. Let’s see if this works. I get -16 times -6, 96, minus 1. That’s 95t, and I also get a plus 6. So this works. -16t². There are two solutions to this. T equals -1, this is -1/16 from here, or t equals 6. Now of those two solutions, one of them doesn’t make sense. One of them happened before the ball was thrown. The t equals -16. So we’ll eliminate that solution. We’re concentrating on t equals 6. That’s when the ball actually strikes ground. So how long was the ball in the air? A; the ball was in the air for 6 seconds. Then B;) what’s its velocity when it strikes the ground? I use my velocity equation and I plug in 6, because that’s the time when it strikes the ground. So v(6) is what I need.

-32 times 6 plus 95. -36 times is -192, plus 95, this is going to give me -97. This is the velocity, -97 it's in feet per second. It makes sense that the velocity is negative because if you look back here, the object’s going to be on its way down. Its velocity should be in the negative direction. But when we’re asked about speed, were not asked about velocity. =

Speed is different from velocity. Speed is the absolute value of velocity. They want to know the magnitude of this answer. They want to know the 97. 97 feet per second is the speed. Just remember v(6) equals -97, that’s the velocity at 6 seconds. The speed at 6 seconds is 97 feet per second, but you can say downward if you like, to indicate the direction. And that’s it.

Tags
differential equations derivatives position velocity acceleration