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

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Answer 1 · 2018-03-14T07:29:41

The time is $=3.28s$

The distance $AB$ is

$AB=sqrt((9-2)^2+(3-1)^2+(5-3)^2)=sqrt(49+4+4)=sqrt(57)m$

Apply the equation of motion

$s=ut+1/2at^2$

The initial velocity is $u=0ms^-1$

The acceleration is $a=7/5ms^-2$

Therefore,

$sqrt57=0+1/2*7/5*t^2$

$t^2=10/7*sqrt57$

$t=sqrt(10/7*sqrt57)=3.28s$

An object is at rest at (2 ,1 ,3 )(2 ,1 ,3 ) and constantly accelerates at a rate of 7/5 m/s^27/5 m/s^2 as it moves to point B. If point B is at (9 ,3 ,5 )(9 ,3 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.