The velocity (mi\operatorname{mi}/hrhr) of a drone flying in the air is given by v(t)=12+4t2v\left(t\right)=12+4t^2 for 0≤t≤40\le t\le4 hours. Let s(0)=0s\left(0\right)=0. Determine s(t)s\left(t\right) for 0≤t≤40\le t\le4.

s ( t ) = 4 t 3 3 + 12 t s\left(t\right)=\frac{4t^3}{3}+12t

The velocity ( mi \operatorname{mi} / h r hr ) of a drone flying in the air is given by v ( t ) = 12 + 4 t 2 v\left(t\right)=12+4t^2 for 0 ≤ t ≤ 4 0\le t\le4 hours. Let s ( 0 ) = 0 s\left(0\right)=0 . Determine s ( t ) s\left(t\right) for 0 ≤ t ≤ 4 0\le t\le4 .

In this problem, we are given the velocity of a drone in miles per hour, and it's given by this function right here, v of t. We're also given this time interval from zero to four, and we're asked to find our position, s of t. Now to solve this type of problem, what we need to do is know the equation we can use to find s of t, and it looks like this. S of t is going to be equal to the initial position, which we can see that's actually given to us as s of zero, plus the integral of our velocity function with respect to x. We need to use x as a dummy variable here since we don't wanna be integrating with respect to t twice, or we don't want our upper bound to be t, and so that's why we're switching this out with x. And our bounds here, well, that's going to go from zero all the way up to t x. See, our initial position there is zero. We're going to t. So we need to determine this function s of t. Now to solve this, well, first off, I'm going to take s of zero, which is our zero right here. I'm going to say that that's just equal to zero plus this integral right here, which is gonna be the integral from zero to t of our velocity function, which I can see is 12 plus four t squared, but I need to replace that t with x. So it's gonna be 12 plus four x squared, and then this whole thing is integrated with respect to x. Now from here, I can go ahead and just solve this integral. So we're going to have the integral of this function here which is going to be 12x using these power rules that I know about we're going to get the integral of 4x squared which is going to be 4x cubed over three, we're adding this on and it's going to be in a or this is going to be bounded from the bounds that we have up here so it's going to be bounded from zero to t. And since I have zero as my bottom bound I can just take t and plug it in for x. So doing this what I'm going to get as a final result is I will get four t cubed over three, and then we're going to get plus 12 t, and that's going to be our result. Now this right here is our position function s of t. And notice this is the reason why we did not want to have this variable in here be t at the start because we had t as this upper bound. So we want s of t to be a function with respect to t, and so that's why we need this dummy variable in here when we integrate. So that's our solution to how you can solve these types of problems. Hope you found this video helpful, and let's move on.

The velocity ( mi ⁡ \operatorname{mi} / h r hr ) of a drone flying in the air is given by v ( t ) = 12 + 4 t 2 v\left(t\right)=12+4t^2 for 0 ≤ t ≤ 4 0\le t\le4 hours. Let s ( 0 ) = 0 s\left(0\right)=0 . Determine s ( t ) s\left(t\right) for 0 ≤ t ≤ 4 0\le t\le4 .

s ( t ) = 4 t 3 3 + 12 t s\left(t\right)=\frac{4t^3}{3}+12t

s ( t ) = 4 t 2 3 + 12 t s\left(t\right)=\frac{4t^2}{3}+12t

s ( t ) = 8 t s\left(t\right)=8t

s ( t ) = 8 t 3 3 + 4 t s\left(t\right)=\frac{8t^3}{3}+4t^{}

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

The velocity ( $mi$ / $h r$ ) of a drone flying in the air is given by $v (t) = 12 + 4 t^{2}$ for $0 is less than or equal to t is less than or equal to 4$ hours. Let $s of 0 equals 0$ .
Determine $s (t)$ for $0 is less than or equal to t is less than or equal to 4$ .

$s (t) = \frac{4 t^{2}}{3} + 12 t$

$s of t equals 8 t$

$s (t) = \frac{4 t^{3}}{3} + 12 t$

$s (t) = \frac{8 t^{3}}{3} + 4 t^{}$

Verified step by step guidance

Step 1: Recognize that the problem involves finding the position function s(t) from the given velocity function v(t). The position function is the integral of the velocity function with respect to time, as velocity is the derivative of position.

Step 2: Write the given velocity function as v(t) = 12 + 4t^2. To find s(t), compute the integral of v(t) with respect to t: ∫v(t) dt = ∫(12 + 4t^2) dt.

Step 3: Break the integral into two parts for simplicity: ∫(12 + 4t^2) dt = ∫12 dt + ∫4t^2 dt. Now, compute each term separately.

Step 4: For the first term, ∫12 dt, the result is 12t (since the integral of a constant is the constant multiplied by the variable of integration). For the second term, ∫4t^2 dt, use the power rule for integration: ∫t^n dt = (t^(n+1))/(n+1). Applying this, ∫4t^2 dt = (4/3)t^3.

Step 5: Combine the results of the two integrals to get the position function: s(t) = (4/3)t^3 + 12t + C, where C is the constant of integration. Use the initial condition s(0) = 0 to solve for C. Substituting t = 0 into s(t), we find C = 0. Thus, the final position function is s(t) = (4/3)t^3 + 12t.

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