Question

Given the velocity and initial position of a body moving along a coordinate line at time t, find the bodys position at time t.

V = 8/pi sin 4t/pi, s(pi^2) = 2
1. s = -2 cos 4t/pi + 3
2. s = -2 cos 4t/pi + 4
3. s = -2 cos 4t/pi + 8
4. s = -2 cos 4t/pi + 4

Answer 1

To find the body's position at time t given the velocity function and initial position, we integrate the velocity function with respect to t and solve for the constant of integration using the initial position. The resulting position function is s = -2*cos(4t/pi) + 4.

Step-by-step explanation:

To find the body's position at time t, we need to integrate the velocity function. In this case, the velocity function is V = 8/pi * sin(4t/pi). We can integrate this function with respect to t to obtain the position function, s. The antiderivative of sin(4t/pi) is -2*cos(4t/pi). Therefore, the position function is s = -2*cos(4t/pi) + C, where C is the constant of integration.

Given that s(pi^2) = 2, we can substitute t = pi^2 into the position function and solve for C. 2 = -2*cos(4(pi^2)/pi) + C. Simplifying this equation gives C = 4.

Therefore, the body's position at time t is s = -2*cos(4t/pi) + 4.

Answer 2

4. s = -2 cos 4t/pi + 4

Step-by-step explanation:

The position is the integral of the velocity.

In this question:

`v(t) = (8)/(\pi)\sin{((4t)/(\pi))}`

So

`s(t) = \int {v(t)} dt`

`s(t) = \int {(8)/(\pi)\sin{((4t)/(\pi))}} dt`

Integrating by substitution:

`u = (4t)/(\pi)`

`du = (4)/(\pi)dt`

`dt = (\pi du)/(4)`

So

`s = \int {(8)/(\pi)sin(u) ((\pi du)/(4))`

`s = 2 \int {sin(u)} du`

`s = -2 cos(u) + C`

Since
`u = (4t)/(\pi)`

`s(t) = -2 \cos{(4t)/(\pi)} + C`

Since s(pi^2) = 2 .

`s(t) = -2 \cos{(4t)/(\pi)} + C`

`2 = -2 \cos{(4\pi^(2))/(\pi)} + C`

`2 = -2cos(4\pi) + C`

The cosine of 4pi is the same as the cosine of 0 = 1. So

`2 = -2 + C`

`C = 4`

So the correct answer is:

4. s = -2 cos 4t/pi + 4