A particle moves along the xxx-axis and its acceleration is given by a(t)=cosπta\left(t\right)=\cos\pi ta(t)=cosπt. Find s(t)s\left(t\right) if s(0)=1s\left(0\right)=1

s ( t ) = π 2 − cos ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)+1}{\pi^2}

A particle moves along the x x x -axis and its acceleration is given by a ( t ) = cos π t a\left(t\right)=\cos\pi t a ( t ) = cos π t . Find s ( t ) s\left(t\right) if s ( 0 ) = 1 s\left(0\right)=1

In the last video, we were going over this problem where we have a particle moving along the x axis, and we were given this acceleration function right here. Now this video asks us here, this problem, asks us to find the position function s of t, if the initial position is one. So in the last video, we learned about how we could find this velocity function right here. This was the problem that we solved. And now we're trying to find our position function, which we can use our velocity function to find that. So if you haven't seen the last video, definitely check that one out, so that way you can figure out how we got this solution right here. But if you've seen it already, let's go ahead and solve this part of the problem. So what we're looking for is our position s of t. And recall that there's a way we can find position. What you wanna do is take your initial position, and you wanna add it to the integral from the initial time to some time t, this variable here, and then you wanna be integrating your velocity function as a function of x. So that's what we're gonna do here. Now we have our initial position s of zero given in the problem. And what I can do is add this to our integral for the initial time, which we know is zero based on this, to some time t of our velocity function as a function of x. Now plugging this in, we're going to have v of x plugged in here, which is going to be one over pi times the sine of pi x, integrated with respect to x. So notice all I really did here was take this t here, and replace it with x. Now next, I'm gonna try to make this integral and this equation look a little more simple. S of zero, well, that's gonna be one. So we'll have one plus, and then we're going to be integrating from zero to t. But what I'm actually gonna do here is take this one over pi that I see and pull it outside of the integral. So I have one over pi times the integral from zero to t of my function sine of pi x. Now, from here, what we need to do is figure out how we can solve this integral that we see. And this is gonna be a little bit tricky since I noticed that we still have this pi x term inside. But recall in the last video, we were able to do this with a u substitution, so that's what I'm gonna do here. I'm gonna take everything inside here, and I'm going to replace it with u. So we can say, and I'm gonna do this in a different color, just so we don't have the same color as this. So we're gonna have u here. So we're gonna say that u is equal to pi x, meaning that the derivative of u would just be pi, if we're taking this derivative with respect to x. So resetting up this integral here, we're gonna have one plus and it's gonna be one over pi times the integral from zero to t, sine of u, and then this is still being integrated with respect to x. Now we can't integrate this, u function with respect to x. These are two different variables. So what I need to do is try and get this whole thing in terms of u. Well, I can do that by relating the du to what we have down here. So what I'm gonna do is take this d x that I see, and I'm gonna multiply it by pi. So pi is a constant, so we'll have pi d x. But I can't just go ahead and multiply the pi and then be done here, because that would change the argument inside of my integral. So what I'm going to do is then divide by pi as well to keep the same argument. Now this pi in the denominator, I'm just gonna go ahead and combine it with this pi here to go ahead and square that. So I can take this pi times d x that I see there, and I can replace it with this d u. So doing so, I'm just gonna take the pi dx here and replace it with du, and now we have everything in terms of u. So how exactly do I find the integral for the sine of u? Well, if I'm looking for the integral for the sine, that's gonna come out to negative cosine. Because we've seen before that integrating sine gives us negative cosine, so the same thing's gonna happen here. So what that means is the integral of sine will give us negative cosine, but now we're integrating this bounded from zero to t. But I can't just take my t and my zero and plug it in for u, because notice that u was not originally what we started with for these bounds. When we originally applied the bounds, that's when we were in terms of x. So what I need to do is take my original function for u here, or the original function with x in it, and put it back in for u. So I can see that we have pi x that we set here. So doing this, we're gonna have one, and that's gonna be plus one over pi squared. That's gonna be multiplied by negative cosine of pi x, and then that's all gonna be bounded from zero to t. So now we have everything set up for putting these bounds in to our original function once we've integrated that, of course. But here's the thing. You might think that we can just ignore this bottom bound here since it's zero, but this is where you have to be careful. Because if we took zero and plugged it in for x there, we would end up just getting the cosine of zero. But the cosine of zero does not come out to zero. The cosine of zero comes out to one. So this is why you actually have to check to see what would happen as you put zero into your function here. Oftentimes, it does come out to zero, especially when you have polynomials, but when you have trig functions, you have to be a bit careful with this. So since I can see it's not gonna come out to zero, I do have to run these bounds properly. So we're gonna have one plus one over pi squared, and then we're going to have everything in here. Now first off, we're going to take my high bound of t and plug it in for x. And I'm gonna do this all in a different color here just so we're able to keep track of things. So we're gonna have t in for x. And so doing this first, we're gonna have the the high bound negative cosine of pi t. And then what I need to do from here is subtract off zero plugged in for my bottom bound. So we're going to have negative cosine of pi times zero, which would just be zero. So this is what we can plug in. Now as it turns out, we can cancel these negative signs, and the cosine of zero just turns out to be one. So this whole thing is gonna simplify to one plus one over pi squared, and it's gonna be multiplied by we have negative cosine pi t plus one. So this is what we get. So as you can see, we now have this set up, and what I'm gonna do from here is take my one over pi squared and distribute it into these brackets. So that's going to give me one plus, and then we're going to have negative cosine pi t all over pi squared, and then we're gonna have plus, and then one times one over pi squared would just be one over pi squared. So here's what we have now. Now from here, I'm gonna go ahead and continue my math up here just because I wanna make sure that I have enough space. So we still have s of t that we're trying to find. This is what we're trying to find down here throughout this problem as well. So our position s of t, what I can actually do is take this one here, and I can replace it with pi squared over pi squared. Now there's a reason that I'm doing this. If I have a pi squared over pi squared here, notice we now have common denominators across the board. So I can combine this into one common fraction, which is going to be pi squared, dealing with this term, that's gonna be minus cosine of pi t, dealing with that term, plus one, dealing with this term, all over pi squared. So this right here would be a function for our position, and that is a simplified solution that we can get to our problem. So we now have our function of position as a function of time, and that right there is the answer. So as you can see, when solving these types of problems, it oftentimes does take a bit of work, since we need to run this integral and we oftentimes have to do u substitutions in order to make the integral work. But as long as you stick with what we already know when it comes to bounding these integrals and doing these u substitutions and dealing with trigonometric functions, you should be good when it comes to solving these more complicated types of problems. So hope you found this video helpful, and let's move on.

A particle moves along the x x x -axis and its acceleration is given by a ( t ) = cos ⁡ π t a\left(t\right)=\cos\pi t a ( t ) = cos π t . Find s ( t ) s\left(t\right) if s ( 0 ) = 1 s\left(0\right)=1

s ( t ) = π 2 − cos ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 − cos ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)}{\pi^2}

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

Struggling with Calculus?

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Multiple Choice

A particle moves along the $x$ -axis and its acceleration is given by $a\left(t\right)=\cos\pi t$ .
Find $s (t)$ if $s of 0 equals 1$

$s (t) = \frac{π^{2} - \cos (π t)}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t) + 1}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t)}{π^{2}}$

$s (t) = \frac{π^{2} - \cos (π t) + 1}{π^{2}}$

Verified step by step guidance

Step 1: Recall that acceleration a(t) is the second derivative of position s(t) with respect to time t. To find s(t), we need to integrate a(t) twice. The given acceleration is a(t) = cos(πt).

Step 2: Perform the first integration to find the velocity v(t). Integrate a(t) = cos(πt) with respect to t. The integral of cos(πt) is (1/π)sin(πt) + C₁, where C₁ is the constant of integration.

Step 3: Perform the second integration to find the position s(t). Integrate v(t) = (1/π)sin(πt) + C₁ with respect to t. The integral of (1/π)sin(πt) is -(1/π²)cos(πt), and the integral of C₁ is C₁t. Add another constant of integration, C₂.

Step 4: Use the initial condition s(0) = 1 to solve for the constants. Substitute t = 0 into s(t) = -(1/π²)cos(πt) + C₁t + C₂. Since cos(0) = 1, this simplifies to s(0) = -(1/π²)(1) + C₂ = 1. Solve for C₂.

Step 5: Combine the results to write the final expression for s(t). Substitute the values of C₁ and C₂ into s(t) = -(1/π²)cos(πt) + C₁t + C₂. Simplify the expression to match the given correct answer, s(t) = (π² - cos(πt) + 1)/π².

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