An object is at rest at $(3 ,0 ,3 )$ and constantly accelerates at a rate of $7/5 m/s^2$ as it moves to point B. If point B is at $(5 ,3 ,5 )$ , 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 · 2017-07-17T16:18:08

The time is $=2.43s$

The distance $AB$ is

$s=sqrt((5-3)^2+(3-0)^2+(5-3)^2)$

$=sqrt(4+9+4)$

$=sqrt17m$

We apply the equation of motion

$v^2=u^2+2as$

The initial velocity is $u=0ms^-1$

The acceleration is $a=7/5ms^-2$

Therefore,

$v^2=0+2*7/5*sqrt17$

$v^2=11.54$

$v=sqrt(11.54)=3.4ms^-1$

To determine the time, we apply the equation of motion

$v=u+at$

$t=v/a$

$t=3.4/(7/5)=2.43s$

An object is at rest at (3 ,0 ,3 )(3 ,0 ,3 ) and constantly accelerates at a rate of 7/5 m/s^27/5 m/s^2 as it moves to point B. If point B is at (5 ,3 ,5 )(5 ,3 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.