Question

An object is at rest at `(8 ,3 ,3 )` and constantly accelerates at a rate of `7/4 ms^-2` as it moves to point B. If point B is at `(6 ,3 ,4 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

First step is to find the distance between the two points: `r~=2.24` `m`

Second step is to find the time taken: `t=1.6` `s`

Explanation:

To find the distance between two points in 3D space we have:

`r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)`

`=sqrt((6-8)^2+(3-3)^2+(4-3)^2)=sqrt((-2)^2+(0)^2+(1)^2)`

Therefore:

`r=sqrt5~=2.24`

Now to find the time taken, we can use:

`d=ut+1/2at^2`

Since the object is initially at rest, the `ut` term goes to `0`.

Rearranging:

`t=sqrt((2d)/a)`

Note that I have used both `d` and `r` because of the way I remember formulae, but we're talking about the same distance - the distance between the points.

`t=sqrt((2xx2.24)/(7/4))= 1.6` `s`