Question

What is the instantaneous velocity of an object with position at time t equal to `f(t)= (te^(t^3-t),t^2e^t)` at `t=1`?

Answer 1

Instantaneous velocity of the object is `3hati+3ehatj` and its absolute value `sqrt(9+9e^2)`

Explanation:

When position of an object at a time is given by

`f(t)=(te^(t^3-t),t^2e^t)`

This means while `x`-coordinate is `te^(t^3-t)`, `y`-coordinate is given by `t^2e^t` and hence, its velocity is given by

`vecv=(dx)/(dt)hati+(dy)/(dt)hatj`,

where `hati` is unit vector along `x`-axis and `hatj` is unit vector along `y`-axis

and its absolute value is `((dx)/(dt))^2+((dy)/(dt))^2`

Hence the velocity is given by

`vecv=(e^(t^3-t)+t(3t^2-1)e^(t^3-t))hati+(t^2e^t+2te^t)hatj`

and velocity at `t=1` is given by

`vecv_(t=1)=(e^(1^3-1)+(3*1^2-1)e^(1^3-1))hati+(1^2e^1+2e^1)hatj`

= `(1+2)hati+(e+2e)hatj`

= `3hati+3ehatj`

and its absolute value `sqrt(9+9e^2)`