Suppose that a particle travels along the x x -axis and its velocity is given by v ( t ) = 2 cos t v\left(t\right)=2\cos t for 0 ≤ t ≤ 2 π 0\le t\le2\pi . Find the particle's displacement on [ 0 , 5 π 4 ] \left\lbrack0,\frac{5\pi}{4}\right\rbrack

this problem, we are told, suppose that a particle travels along the x axis and its velocity is given by this function right here for this time interval. We're asked to find the particle's displacement on the interval from zero to five pi over four, which that does fall within the time frame that we're given here, so we can go ahead and find this displacement. Now recall that when we're finding a displacement and we're given a velocity, all we really need to do is integrate the velocity function based on our interval. So to find our displacement delta s, that's going to be the integral from zero to five pi over four of our velocity function, which we can see here is two cosine of t, and that's integrated with respect to t. So this is all we're really doing to find our displacement. Now what I'm first going to do is figure out what this integral is here, and I can take that two and bring it outside the integral there. So we're gonna have two times the integral from zero to five pi over four, and that's gonna be the integral of cosine of t with respect to t. Now the integral of cosine, well that's just gonna be sine the anti derivative, so we'll have two times the sine of t, and this whole function here is going to be bounded from zero to five pi over four. Now the sine of zero is just going to be zero, so I only really need to worry about plugging my top bound in right here. So if I take my five pi over four and plug it in for t, what we're gonna end up getting is two times the sine of five pi over four. Now for five pi over four the the sign here if you go ahead and look at a unit circle where you try this on your calculator you should get a result of negative square root two over two. So you'll get two times negative square root two over two, in which case we can cancel this two with that two. So our final result is only going to be negative square root of two. So this right here is the solution to our problem, and that is our displacement. Now notice here that we did get a negative result for the displacement, but that can happen when you're dealing with these problems because we're called the displacement is just a change in position. So let's say, for example, I were to move, say, three meters to the right, but then all of a sudden I change direction, and then I moved three plus the square root of two meters to the left. Well, we could say that the direction to the right was positive, and the direction to the left was negative. So the total distance that we traveled right here, our total displacement in this case, would just be negative square root of two since we moved backwards. So it is possible to get these negative results like we saw down here. That's just something to keep in mind when dealing with these types of problems. But hope you found this video helpful. That's our result. And let me know if you have any questions.

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v of t equals 2 cosine t$ for $0 is less than or equal to t is less than or equal to 2 pi$ .
Find the particle's displacement on $open bracket 0 comma 5 pi over 4 close bracket$

$- \sqrt{2}$

$- \frac{\sqrt{2}}{2}$

$- 2 - \sqrt{2}$

$2 + \sqrt{2}$

Verified step by step guidance

Step 1: Understand the problem. The displacement of a particle is the integral of its velocity function over the given time interval. Here, the velocity function is v(t) = 2cos(t), and the interval is [0, 5π/4].

Step 2: Write the integral for displacement. The displacement is given by the definite integral: ∫[0, 5π/4] 2cos(t) dt.

Step 3: Solve the integral. The antiderivative of 2cos(t) is 2sin(t). Use this to evaluate the definite integral: ∫[0, 5π/4] 2cos(t) dt = [2sin(t)] evaluated from t = 0 to t = 5π/4.

Step 4: Apply the limits of integration. Substitute the upper limit (t = 5π/4) and the lower limit (t = 0) into the antiderivative: 2sin(5π/4) - 2sin(0).

Step 5: Simplify the result. Use the unit circle to find sin(5π/4) and sin(0). sin(5π/4) corresponds to the sine value at 225 degrees, which is -√2/2, and sin(0) is 0. Substitute these values into the expression to find the displacement.