Question

The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.

Answer 1

Answer:
`x=(2)/(3)t^3+3t+1`

Step-by-step explanation:

Given

velocity of object is given by

`v(t)=3+2t^2`

and we know change of position w.r.t time is velocity

`\Rightarrow (dx)/(dt)=v`

`\Rightarrow (dx)/(dt)=3+2t^2`

`\Rightarrow dx=(3+2t^2)dt`

Integrating both sides we get

`\Rightarrow \int_(1)^(x)dx=\int_(0)^(t)(3+2t^2)dt`

`\Rightarrow x\mid _(1)^(x)=(3t+(2)/(3)t^3)\mid _(0)^(t)`

`\Rightarrow x-1=3(t-0)+(2)/(3)(t^3-0)`

`\Rightarrow x=(2)/(3)t^3+3t+1`