An object is at rest at $(8 ,2 ,5 )$ and constantly accelerates at a rate of $7/4 m/s^2$ as it moves to point B. If point B is at $(6 ,5 ,1 )$ , 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 · 2016-05-14T07:22:33

time $t=2.480821248" "$ seconds

Given object at rest
$v_0=0" "$ initial velocity

distance traveled by the object
$s=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)$
$s=sqrt((6-8)^2+(5-2)^2+(1-5)^2)$
$s=sqrt(4+9+16)$
$s=sqrt(29)$

the formula to compute t

$s=v_0t+1/2at^2$
$sqrt(29)=(0)*t+1/2(7/4)t^2$
$t^2=8/7sqrt29$
$t=(8/7sqrt29)^(1/2)$

$t=2.480821248" "$ seconds

God bless....I hope the explanation useful

An object is at rest at (8 ,2 ,5 )(8 ,2 ,5 ) and constantly accelerates at a rate of 7/4 m/s^27/4 m/s^2 as it moves to point B. If point B is at (6 ,5 ,1 )(6 ,5 ,1 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.