Question

What is the instantaneous velocity of an object with position at time t equal to `f(t)= (t^2-8,tsqrt(t^2+3t-4))` at `t=2`?

Answer 1

Velocity vector `v(t)=(2t, (t(2t+3))/(2sqrt(t^2+3t-4))+sqrt(t^2+2t-4))`, and so,

`v(2)=(4, 7/sqrt 6 + sqrt 6)=(4, 5.31)`, nearly.

Explanation:

The position vector `r=r(x, y)=(x(t), y(t))=(t^2-8, t sqrt(t^2+3t-4))=f(t).`

So, the velocity vector `v(t)=(x'(t), y'(t))`

`=f'(t)=(2t, (t(2t+3))/(2sqrt(t^2+3t-4))+sqrt(t^2+2t-4))`

So, `v(2)=(4, 7/sqrt 6 + sqrt 6)`.