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 ,4 ,7 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

It will take the object `5` seconds to reach point B.

Explanation:

You can use the equation
`r = v \Delta t + 1/2 a \Delta t^2`
where `r` is the separation between the two points, `v` is the initial velocity (here `0`, as at rest), `a` is acceleration and `\Delta t` is the elapsed time (which is what you want to find).

The distance between the two points is
`(3,4,7) - (2,1,6) = (3-2, 4-1, 7-6) = (1,3,1)`

r = || (1,3,1) || = `\sqrt(1^2 + 3^2 + 1^2) = \sqrt{11} = 3.3166 \text{m}`

Substitute `r = 3.3166`, `a = 1/4` and `v=0` into the equation given above
`3.3166 = 0 + 1/2 1/4 \Delta t^2` Rearrange for `\Delta t`
`\Delta t = \sqrt{(8)(3.3166)}`
`\Delta t = 5.15 \text{s}`

Round to however many decimal places are requested, or to significant figures, of which here there is one, so `5s`.