- #1

fishface1959

- 3

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The s^2. s^3 etc... are seconds

## Homework Statement

A particle is moving in one-dimension, and its acceleration in m/s[tex]^{2}[/tex]as function of time can be written as:

a(t) = 3.0m/s^3 t + 6.0m/s[tex]^{2}[/tex]sin(0.5/s t)

Assume: v(0) = 3.5m/s and x(0) = 5.0m

Derive the function of velocity v(t)

Calculate the instantaneous velocity v(t) when t = 2.0s

**The attempt at a solution**

This is what I come up with for v(t)

v(t) =integral a(t) + Vi

= 3.5m/s + (3.0m/s^3[tex]\int[/tex] t) + (6.0m/s^2 integral sin(u)du)

= 3.5m/s + 1.5m/s^3 t^2 - (6.0m/s^2)cos(.5/s t)(2)

The answer states:

v(t) = 15.5 m/s + 1.5m/s^3 t^2 - 12.0 m/s cos(.5/s t)

v(2s) = 15.02m/s

I do not know where the 15.5 comes from. I realize that at t=0, cos(0) = 1 , I'm not sure how or if this plays into the V(t) equation.

Using the equation to the answer v(2s) answer I come up with V(2s) = 9.5 - 3.5=6 (approx), not what the answer is supposed to be.

Could someone please tell me what I am missing and redirect me to what I need to learn.

Thank you