Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (e^(t-t^2),2t-te^t)` at `t=-1`?

Answer 1

`v=3.5791`(Length unit/unit of time)
and it's direction `theta=253.7756`degrees

Explanation:

`t` can't be negative as the time can't be negative but the following answer demonstrates the way to solve this kind of equations for `t=1` instead of `t=-1` and I will put the answer on double check.

This function is a displacement function for a particle moving in the Cartesian plane:
The function is divided into two parts `x,y` which are both parametric equations
so You can differentiate each part alone to get the velocity-time function
`f'(t)=((1-2t)e^(t-t^2),2-e^t-te^t)`

and by substituting with the value `t=-1` You get:

`f'(1)=((-1)e^0,2-e^1-(1)e^1)`
and by simplification You get:
`f'(-1)=(-1,2(1-e))`
so the magnitude of the velocity of the particle at that moment will be:
`v`=`sqrt((x^o)^2+(y^0)^2)`

and it's direction will be `tantheta=(y^0)/(x^0)`

And by the substituting, You get `v=3.5791`(Length unit/unit of time)
and `theta=253.775degrees`

I hope this was helpful.