An object is at rest at $(7 ,8 ,4 )$ and constantly accelerates at a rate of $5/4 m/s^2$ as it moves to point B. If point B is at $(9 ,5 ,3 )$ , 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 · 2017-01-03T06:51:19

The answer is $=2.45s$

The distance betwwen A and B is

$=sqrt((9-7)^2+(5-8)^2+(3-4)^2)$

$=sqrt(4+9+1)$

$=sqrt14m$

We use the equation

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

$u=0$

So,

$sqrt14=1/2*5/4*t^2$

$t^2=8/5sqrt14=5.99s^2$

$t=2.45s$

An object is at rest at (7 ,8 ,4 )(7 ,8 ,4 ) and constantly accelerates at a rate of 5/4 m/s^25/4 m/s^2 as it moves to point B. If point B is at (9 ,5 ,3 )(9 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.