An object is at rest at $(4 ,5 ,1 )$ and constantly accelerates at a rate of $4/3 m/s^2$ 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 · 2016-06-18T16:13:16

It takes $3.14$ s.

The equation of motion for a constant accelerating object is

$s=1/2at^2$ where $s$ is the space traveled, $a$ is the acceleration and $t$ is the time.

We are interested in the time that is

$t=sqrt((2s)/a)$ .

The acceleration is given, then we need the space.
The distance between two point in space $p_1(x_1, y_1, z_1), p_2(x_2, y_2, z_2)$ is given by

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

For us

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

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

$=sqrt(9+9+25)=sqrt(43)\approx6.56$ .

This is the distance that the object has to travel, so we can write
$s=6.56$ m. We now have all the ingredients

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

$=sqrt((2*6.56" m")/(4/3"m"/"s"^2))$

$=sqrt((13.12" m")/("m")*3/4 "s"^2)$

$=sqrt(9.84" s"^2)$

$=sqrt(9.84" s"^2)\approx3.14" s"$ .

An object is at rest at (4 ,5 ,1 )(4 ,5 ,1 ) and constantly accelerates at a rate of 4/3 m/s^24/3 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.