An object is at rest at $(2 ,8 ,6 )$ and constantly accelerates at a rate of $5/3 m/s^2$ as it moves to point B. If point B is at $(7 ,5 ,3 )$ , 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 · 2017-05-07T06:02:13

The time is $=2.81s$

The distance between the points $A=(x_A,y_A,z_A)$ and the point $B=(x_B,y_B,z_B)$ is

$AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)$

$d=AB= sqrt((7-2)^2+(5-8)^2+(3-6)^2)$

$=sqrt(5^2+3^2+3^2)$

$=sqrt(25+9+9)$

$=sqrt43$

$=6.56$

We apply the equation of motion

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

$u=0$

so,

$d=1/2at^2$

$t^2=(2d)/a=(2*6.56)/(5/3)$

$=7.87$

$t=sqrt(7.87)=2.81s$

An object is at rest at (2 ,8 ,6 )(2 ,8 ,6 ) and constantly accelerates at a rate of 5/3 m/s^25/3 m/s^2 as it moves to point B. If point B is at (7 ,5 ,3 )(7 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.