Question

An object is at rest at `(2 ,1 ,6 )` and constantly accelerates at a rate of `1/4 m/s^2` as it moves to point B. If point B is at `(3 ,2 ,7 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

`t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25) = 3.72` `s`

Explanation:

In a 3-dimensional coordinate system, the distance between two points is given by:

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

`r=sqrt((3-2)^2+(2-1)^2+(7-6)^2) =sqrt3 = 1.73` `m`

We can use the equation of motion:

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

Since the initial velocity `u=0`, we can ignore the first term, and rearrange `s= 1/2 at^2` to make `t` the subject.

Note that the term 'r' I was using in the distance calculation above is the same as the distance 's' that the object moves. I don't want the symbol change to be confusing.

`t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25)`

(I used 0.25 rather than 1/4 for ease of writing, but of course it means the same thing)

Therefore `t = 3.72` `s`.