Question

An object is at rest at `(2 ,9 ,8 )` and constantly accelerates at a rate of `1/5 m/s` as it moves to point B. If point B is at `(6 ,2 ,7 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

`t = (66)^(1/4)sqrt(10) s`

Explanation:

I'm assuming the object travels in a straight line. Distance between the two points given by:

`Deltar = sqrt((Deltax)^2+(Deltay)^2+(Deltaz)^2)`

`= sqrt((6-2)^2+(2-9)^2+(7-8)^2) = sqrt(4^2+7^2+1^2) = sqrt(66)`

The object is travelling with constant acceleration along this path, there are no external forces acting on it so equation of motion is:

`(d^2r)/(dt^2) = 1/5`

Integrating gives

`(dr)/(dt) = 1/5t + C_1`

This may be reminiscent of `v = u + at` because, well, it's the same thing. `C_1` denotes initial speed here. As we start from rest, in this case it is equal to zero.

Integrating again gives

`r(t) = 1/10t^2 + C_2`

We know that `r(0) = 0` so `implies C_2 = 0`

`therefore r(t) = 1/10t^2`

The time at which `r = sqrt(66)` is then found by:

`sqrt(66) = 1/10t^2 implies t^2 = 10*sqrt(66)`

Hence `t = (66)^(1/4)sqrt(10) s`

(we discard the negative solution as we are working with times).