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

Answer:

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

Step-by-step explanation:

The position is the integral of the velocity.

In this question:

$v(t) = \frac{8}{\pi}\sin{(\frac{4t}{\pi})}$

So

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

$s(t) = \int {\frac{8}{\pi}\sin{(\frac{4t}{\pi})}} dt$

Integrating by substitution:

$u = \frac{4t}{\pi}$

$du = \frac{4}{\pi}dt$

$dt = \frac{\pi du}{4}$

So

$s = \int {\frac{8}{\pi}\sin{u} (\frac{\pi du}{4})$

$s = 2 \int {\sin{u}} du$

$s = -2 \cos{u} + C$

Since $u = \frac{4t}{\pi}$

$s(t) = -2 \cos{\frac{4t}{\pi}} + C$

Since s(pi^2) = 2 .

$s(t) = -2 \cos{\frac{4t}{\pi}} + C$

$2 = -2 \cos{\frac{4\pi^{2}}{\pi}} + C$

$2 = -2\cos{4\pi} + C$

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