An object is at rest at $(8 ,3 ,8 )$ and constantly accelerates at a rate of $7/4 m/s^2$ as it moves to point B. If point B is at $(6 ,3 ,4 )$ , 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 · 2017-03-18T12:57:11

The time is $=2.26s$

We are going to apply the following equation of motion :

$s=u_0t+1/2at^2$

$u_0=0$

So,

$s=1/2at^2$

$a=7/4ms^-2$

The distance between 2 points $A=(x_1,y_1,z_1)$ and $B=(x_2.y_2.z_2)$ is

$=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

Here,

$A=(8,3,8)$ and $B=(6,3,4)$

So,

$s=sqrt((6-8)^2+(3-3)^2+(4-8)^2)$

$=sqrt(4+0+16)$

$=sqrt20$

From the equation of motion,

$t^2=(2s)/a$

$t^2=(2*sqrt20)/(7/4)=5.11$

$t=sqrt5.11=2.26s$

An object is at rest at (8 ,3 ,8 )(8 ,3 ,8 ) 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 ,3 ,4 )(6 ,3 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.