Question

An object is at rest at `(7 ,8 ,4 )` and constantly accelerates at a rate of `5/4 m/s^2` as it moves to point B. If point B is at `(1 ,5 ,3 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

It takes 3.29 seconds.

Explanation:

First of all we need to know how much distance the objects has to travel. The distance between two points is given by the Pitagora's theorem in 3 dimensions

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

in our case

`d=sqrt((7-1)^2+(8-5)^2+(4-3)^2)=sqrt(36+9+1)=sqrt(46)\approx6.78m`

Now that we know that our body will travel for 6.78 m we can use the equation of motion for an accelerating body that is

`x=1/2at^2`

where `x` is the traveled distance, for us 6.78, `a` is the acceleration, for us `5/4m/s^2` and `t` is the time that we want to find. I plug the values in the equation

`6.78=1/2*5/4*t^2`
`6.78=5/8t^2`

I am interested in `t` so I multiply both sides for `8/5` and do the square root

`sqrt(6.78*8/5)=t`
`t=sqrt(10.848)\approx 3.29` s.