An object is at rest at $(4 ,5 ,1 )$ and constantly accelerates at a rate of $5/3 m/s^2$ as it moves to point B. If point B is at $(7 ,2 ,6 )$ , 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 · 2017-05-03T05:16:40

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

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

$d=AB= sqrt((7-4)^2+(2-5)^2+(6-1)^2)$

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

$=sqrt(9+9+25)$

$=sqrt43$

$=6.56$

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 (4 ,5 ,1 )(4 ,5 ,1 ) 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 ,2 ,6 )(7 ,2 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.