An object is at rest at $(1 ,7 ,2 )$ and constantly accelerates at a rate of $1 m/s^2$ as it moves to point B. If point B is at $(3 ,7 ,4 )$ , 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 · 2016-12-22T06:34:00

The answer is $=2.4s$

The distance between 2 points A $(x_A,y_A,z_A)$ and

B $(x_B,y_B, z_B)$ is

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

Here we have, A $(1,7,2)$ and B $(3,7,4)$

so,

$d=sqrt((3-1)^2+(7-7)^2+(4-2)^2)$

$=sqrt(4+4)=sqrt8m$

We use the equation

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

As, the object is initially at rest $u=0$

$a=1ms^(-2)$

Then,

$sqrt8=1/2*1*t^2$

$t^2=2sqrt8$

$t=sqrt(2sqrt8)=2.4s$

An object is at rest at (1 ,7 ,2 )(1 ,7 ,2 ) and constantly accelerates at a rate of 1 m/s^21 m/s^2 as it moves to point B. If point B is at (3 ,7 ,4 )(3 ,7 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.