Question

If anyone could help that would be great please provide a full explanation

Answer 1

1. v(t)=2sin(2πt)

2.
`s(t) =- (1)/(\pi) cos(2\pi t ) -(1)/(\\\pi ^(2) )`

3. t =
`(1)/(2)`

4. - 0.41994

5. 0.41994

Velocity Function v(t)

The velocity is the integral of the acceleration with respect to time. Given the acceleration function a(t)=
`4cos(2\pi t)}`, we can find the velocity function v(t) by integrating a(t) with respect to t:

`v(t) = \int\limits {a(t)} \, dt`

`v(t) = \int\limits {4cos(2\pi t)} \, dt`

Integrating, we get:

v(t)=2sin(2πt)+C

Since the initial velocity is given as v=0 at t=0, we can find the constant

C=0, and the velocity function is:

v(t)=2sin(2πt)

Position Function s(t):

The position is the integral of the velocity with respect to time. Given the velocity function v(t), we can find the position function s(t) by integrating v(t) with respect to t:

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

`s(t) = \int\limits {2sin(2\pi t)\, dt`

Integrating, we get:

`s(t) =- (1)/(\pi) cos(2\pi t ) +C_(2)`

Given the initial position s(0) =
`-(1)/(\\\pi ^(2) )`

`-(1)/(\\\pi ^(2) ) =- (1)/(\pi) cos(2\pi t ) +C_(2)`

So,
`C_(2) =-(1)/(\\\pi ^(2) )` and the position function is:
`s(t) =- (1)/(\pi) cos(2\pi t ) -(1)/(\\\pi ^(2) )`

First Positive Time t when the Particle is Stationary:

The particle is stationary when the velocity v(t) is equal to zero. Set

v(t)=0 and solve for t:

v(t)=2sin(2πt)

`v(0)=2sin(2\pi t)`

This occurs when

2πt=kπ, where k is an integer. The first positive solution is t =
`(1)/(2)`

Displacement after One Second:

To find the displacement after one second, evaluate the position function

s(t) at t=1

`s(1) =- (1)/(\pi) cos(2\pi ) -(1)/(\\\pi ^(2) )` = - 0.41994

Total Distance Traveled after One Second:

To find the total distance traveled, consider the absolute value of the displacement, since distance is a scalar. The total distance traveled after one second is ∣s(1)∣

`|s(1) =- (1)/(\pi) cos(2\pi ) -(1)/(\\\pi ^(2) )|` = 0.41994