Question

Then atwhat instant of time will its velocity and accelerationvectors be perpendicular to each other?

Answer 1

Given data:

* The x-coordinate of the position at time t is,

`x=t^2`

* The y-coordinate of the position at time t is,

`y=t^3-2t`

Solution:

The position vector at time t is,

`\vec{r}=t^2i+(t^3-2t)j`

The velocity vector at time t is,

`\vec{v}=\frac{\vec{dr}}{dt}`

Substituting the known values,

`\begin{gathered} \vec{v}=(d)/(dt)(t^2i+(t^3-2t)j) \\ \vec{v}=2ti+(3t^2-2)j \end{gathered}`

The acceleration vector at time t is,

`\vec{a}=(d)/(dt)(\vec{v})`

Substituting the known values,

`\vec{a}=(d)/(\differentialDt t)(2ti+(3t^2-2)j)`