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

v ( t ) = 1 π sin ⁡ ( π t ) v\left(t\right)=\frac{1}{\pi}\sin\left(\pi t\right)

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

particle moving along the x axis is accelerated and given by this function right here, a of t equals cosine of pi t. Find the velocity function as a function of time if the initial velocity, b of zero, is zero. Okay, so how do we go about solving this? Well, what we need to recognize here is that we are given an acceleration function, and we're looking for a function of velocity. So I know that integrating my acceleration function would give me velocity, based on the rules we've discussed. And specifically, if I'm looking to solve for velocity, v of t is going to be equal to the initial velocity, that's going to be plus the integral from our initial time to some time t, of our acceleration function as a function of x. This is the equation we talked about where we need this dummy variable here in order to integrate this acceleration with this upper bound of t. So doing this, let's go ahead and try finding the velocity function. So the velocity function is going to be the initial velocity b of zero plus the integral from zero, our initial time, to t of our acceleration function as a function of x. You can see we're given cosine pi t here, so that means we're gonna have cosine of pi x, and that's gonna be integrated with respect to x. So how do I go about solving things from here? Well, what I need to do is I need to go ahead and figure out how I can integrate this portion of the function. Now I know that v of zero, that's just gonna be zero. That's told right here. So I really only need to focus on this integral right there. So what we'll do is we'll have this integral from zero to t of the cosine of pi x integrated with respect to x. Now notice here that we have this function that is within a function. So we have pi x that is inside this cosine operation. And what I'm gonna do is a variable substitution here, so you can clearly see all the steps to solving this problem. So pi x I'm going to set equal to the variable u. So we can see that u is pi x, meaning that the derivative of u, the derivative of pi x with respect to x, would just be pi. Then I need to show this dx here, so we know that we took this derivative with respect to x. So this right here would be my du. Now what I can do is I can say, well, since I've gone ahead and done this substitution, then my whole integral is gonna go from zero to t, then we're gonna have the cosine of u, because we went ahead and substituted u inside of this cosine. Now, on the outside here, notice I still have a dx, and I can't integrate something with the, of u, some function of u with respect to x. But what I can do is I can get this somehow match the du we have here. And I can recognize that I can take this dx here and multiply it by pi. Now I can't just multiply this whole thing by pi because that would change the argument. But if I once again divide by pi, notice this is going to cancel the pi's there, leaving me with just the dx left over. So this is still the same argument we had before. And I can bring this pi in the denominator out front of the integral. So we're going to have one over pi, it's going to be this integral of zero to t, we'll have the cosine of u, and pi dx, well that's the same thing as du we set up here. So that's going to be integrated with respect to u. Now we can go ahead and integrate this function. So we're gonna have one over pi, and then integrating the cosine of u, well the integral of cosine should give us sine. So that means we're going to have the sine of u, and this whole thing is going to be bounded from zero to t. Now I can't deal with the bounds just yet, because notice that if I plugged in this t and the zero, these were originally bound set when we had this as a function of x. But since we now have a function of u, well, I need to take my u from before and plug it back in. So doing this, we're going to have one over pi, then we're going to have the sine of, and then in this case, our u was pi x, so we'll have the sine of pi x, then this whole thing is bounded from zero to t. Now at this point, we need to think about what's gonna happen if we plug in these bounds. Well, notice my lower bound is zero. Now you wanna be careful when dealing with trig functions, because sometimes when you plug zero into a trig function, it's not always gonna give you zero as a result. However, if we plug a zero in here, we'd have zero times pi, which is zero, and the sine of zero actually does come out to zero in this case. So the zero is gonna set this whole trig function to nothing. So we only need to worry about this upper bound in this example. So doing this, we're going to have one over pi times the sine, and then I can just take that x and replace it with my upper bound t. So the sine of pi t. And this right here would be our velocity function as a function of time, and that is the solution to this problem. So we have one over pi times the sine of pi t, which you could also write as sine of pi t over pi if you would prefer. But either way, that's how you can get our velocity function, and that's how we can find the solution. So hope you found this video helpful, and let's move on.

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

v ( t ) = 1 π sin ⁡ ( π t ) v\left(t\right)=\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = − π sin ⁡ ( π t ) v\left(t\right)=-\pi\sin\left(\pi t\right)

v ( t ) = − 1 π sin ⁡ ( π t ) v\left(t\right)=-\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = 1 π cos ⁡ ( π t ) − 1 v\left(t\right)=\frac{1}{\pi}\cos\left(\pi t\right)-1

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

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

$v (t) = \frac{1}{π} \sin (π t)$

$v (t) = - π \sin (π t)$

$v (t) = - \frac{1}{π} \sin (π t)$

$v (t) = \frac{1}{π} \cos (π t) - 1$

Verified step by step guidance

Step 1: Recall that velocity v(t) is the integral of acceleration a(t) with respect to time t. The acceleration is given as a(t) = cos(πt). To find v(t), we need to compute the integral of cos(πt) with respect to t.

Step 2: Set up the integral: v(t) = ∫cos(πt) dt. Use the substitution method to simplify the integral. Let u = πt, so du = π dt. Rewrite the integral in terms of u: ∫cos(u) * (1/π) du.

Step 3: Integrate cos(u) with respect to u. The integral of cos(u) is sin(u). Therefore, ∫cos(u) du = sin(u). Substitute back u = πt to get (1/π)sin(πt).

Step 4: Add the constant of integration C to the result: v(t) = (1/π)sin(πt) + C. To determine the value of C, use the initial condition v(0) = 0.

Step 5: Substitute t = 0 into v(t): v(0) = (1/π)sin(π*0) + C = 0. Since sin(0) = 0, this simplifies to C = 0. Therefore, the final expression for v(t) is v(t) = (1/π)sin(πt).

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