Question

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

Answer 1

`(-5/e^6, 4-3e^2)`

Explanation:

By definition, velocity is the rate of change in position and may hence be found by differentiating the position function with respect to time. This will require application of the product rule and the result is

`therefore v(t)=dx/dt=f'(t)`

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

`v(2)=(-6/(e^6)+1/(e^6),4-2e^2-e^2)`

`=(-5/e^6, 4-3e^2)`

So if this is a normal 2 dimensional vector space with standard notations and units, then the instantaneous velocity at `t=2` may be represented in terms of the standard basis unit vectors as

`v(2)=-5/e^6 hati + (4-3e^2) hatj` `m//s`