Question

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

Answer 1

`t=3,44" s"`

Explanation:

`"distance between two points:"`
`Delta x=3-4=-1`
`Delta y=1-7=-6`
`Delta z=6-1=5`
`s=sqrt((Delta x)^2+(Delta y)^2+(Delta z)^2)`
`s=sqrt((-1)^2+(-6)^2+5^2)`
`s=sqrt(1+36+25)" "s=sqrt62" m"`
`s=7,87 " m"`

`s=1/2*a t^2`
`7,87=1/cancel(2)*cancel(4)/3*t^2`
`t^2=(7,87*3)/2`
`t^2=11,81`
`t=sqrt (11,81)`
`t=±3,44`
`t=cancel(-3,44)" there is no negative time"`
`t=3,44" s"`

Answer 2

`t=sqrt(3sqrt(31/2))=3.4367`

Explanation:

Given Location of an object,
Rest position `A(4,7,1)`
New Location `B(3,1,6)`
Acceleration, `a=4/3m/s^2`
Find the time it take to move from `A -> B`
Using the distance formula we find the distance from `A->B`
`bar(AB) =r= sqrt((4-3)^2+(7-1)^2 +(1-6)^2) = sqrt(62)`
Now we need to use the kinematic equation of motion that uses
`t, a, r` as variable.
`F(r,t;v_o,a) => r = v_ot +1/2at^2`; `v_o = 0, r=sqrt(62); a=4/3m/s^2`
`sqrt(62) = 1/2*4/3t^2`
`t=sqrt(3sqrt(31/2))=3.4367`