Question

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

Answer 1

We have the parametric curve which define the movement of your material point `M`

We need to derive once to have the parametric curve which define the velocity of your material point `M` let's do that :

`x(t) = tsqrt(t+2)`
`y(t) = t^2-2t`

`vec(OM)(t) = (x(t),y(t))`

`dx/dt= dot(x) = v_x(t) = sqrt(t+2)+t/(2sqrt(t+2)) = (3t+4)/(2sqrt(t+2))`

`dy/dt = dot(y) = v_y(t) = 2t-2`

`(dvec(OM))/dt = vec(v)(t) = (v_x(t),v_y(t)) = ((3t+4)/(2sqrt(t+2)),2t-2)`

`vec(v)(1) = (7/(2sqrt(3)),0)`

Your material point `M` doesn't have a movement relative to the y-axis, so the norm is

`v(1) = 7/(2sqrt(3))`

Answer 2

If you calculate the function of distance in time `f_d(t)`from the given function of position in time `f(t)`, you can derive that the instantaneous velocity of the object when `t=1` is `7/4`.

Explanation:

• The instantaneous velocity of an object is the derivative of distance as a function of time.

• The distance traveled by the object can be taken as its the distance between (0,0) and the current position of the object.

• If distance as a function of time is `y=f_d(t)`, velocity is denoted `v`, the function of position in time is `f(t)` then:
  `v=(dt)/(dy)`
  `f_d(t)= sqrt((t^2-2t)^2+(tsqrt(t+2))^2)`
  `f_d(t)=sqrt((t^4-4t^3+4t^2)+(t^3+2t^2))`
  `f_d(t)=sqrt(t^4-3t^3+6t^2)=(t^4-3t^3+6t^2)^(1/2)`

• `v=(dt)/(dy)=d((t^4-3t^3+6t^2)^(1/2))/(dy)`
  `v=1/2[(t^4-3t^3+6t^2)^(-1/2)][4t^3-9t^2+12t]`
  `v=(4t^3-9t^2+12t)/(2sqrt(t^4-3t^3+6t^2))`
  `v=(4t^2-9t+12)/(2sqrt(t^2-3t+6))`

• When `t=1`
  `v=(4-9+12)/(2sqrt(1-3+6))`
  `v=7/4`

I think?