Question

An object is at rest at `(7 ,4 ,2 )` and constantly accelerates at a rate of `4/5 m/s` as it moves to point B. If point B is at `(3 ,1 ,9 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

Approximately 4.6 seconds

Explanation:

Assuming that the rate of acceleration is `4/5 m/(sec^2)` (that the "squared" was left off the "seconds" of the acceleration.)

Distance between `(7,4,2)` and `(3,1,9)`
`color(white)("XXX")d=sqrt((7-3)^2+(4-1)^2+(2-9)^2)`
`color(white)("XXXX")=sqrt(74)`

The relationship between distance, acceleration (from zero), and time is given by the formula:
`color(white)("XXX")d=1/2a*t^2`

Therefore
`color(white)("XXX")sqrt(74)m=1/(cancel(2)) * cancel(4)^2/5 m/cancel(sec^2) * (t^2 cancel(sec^2))color(white)("/x")`

`color(white)("XXX")t^2=(5sqrt(74))/2`

(using a calculator)
`color(white)("XXX")t=sqrt((5sqrt(74))/2)~~4.637436`