Question

An object is at rest at `(7 ,4 ,2 )` and constantly accelerates at a rate of `4/3 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

`t_F ~~ 3.592" "` to 3 decimal places

Explanation:

`color(red)("Note that I have shown how to deal with units")`

`color(green)("Tip:- leave the converting to decimal until the end")`

`color(blue)("Determine the distance of travel")`

Let distance be `d`
Let initial time be `t_0` seconds
Let time at any instant `i` be `t_i` seconds
Let final time be `t_F` seconds

Given: acceleration `-> 4/3color(white)(.)color(red)( m/s^2)`

Let point 1 be `P_1->(x_1,y_1,z_1)->(7,4,2)`
Let point 2 be `P_2->(x_2,y_2,z_2)->(3,1,9)`

Using Pythagoras

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

`d=sqrt((3-7)^2+(1-4)^2+(9-2)^2)`

`d=sqrt(16+9+49)`

`d=sqrt(74)" "` metres`" "`(74 is a product of 2 primes)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine time of travel")`

Velocity at `t_0=0color(white)(.) m/s`

Velocity at `t_i = 4/3 xx t_i color(green)(" units"-> m/s^sxxs color(blue)(->mxxs/s^2=m/s))`

Mean velocity `1/2xx4/3xxt_F color(white)(.)color(blue)(m/s)`

Thus distance is mean velocity times time giving

`" "d=1/2xx4/3xxt_Fxxt_F`

`" "d= 4/6(t_F)^2 color(green)(" units"-> m/sxxs ->color(blue)(mxxs/s=m`

Multiply both sides by `6/4`

`" "d6/4=4/6xx6/4(t_F)^2`

But `4/6xx6/4=1`

`" "=>d6/4=(t_F)^2`

Thus

`" "t_F=sqrt(6/4)xxsqrt(sqrt(74))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider `sqrt(6/4) -> sqrt((6)/(2^2))=1/2sqrt(6)`

Consider `sqrt(sqrt(74)) -> root(4)(74)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Putting it all together")`

`" "t_F=1/2sqrt(6)xxroot(4)(74)`

By calculator: `t_F ~~ 3.592" "` to 3 decimal places