# An object is at rest at (4 ,5 ,8 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (7 ,9 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

May 9, 2016

Find the distance, define the motion and from the equation of motion you can find the time. Answer is:

$t = 3.423$ $s$

#### Explanation:

Firstly, you have to find the distance. The Cartesian distance in 3D environments is:

Δs=sqrt(Δx^2+Δy^2+Δz^2)

Assuming the coordinates are in form of $\left(x , y , z\right)$

Δs=sqrt((4-7)^2+(5-9)^2+(8-2)^2)

Δs=7.81 $m$

The motion is acceleration. Therefore:

$s = {s}_{0} + {u}_{0} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$

The object starts still $\left({u}_{0} = 0\right)$ and the distance is Δs=s-s_0

$s - {s}_{0} = {u}_{0} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$

Δs=u_0*t+1/2*a*t^2

$7.81 = 0 \cdot t + \frac{1}{2} \cdot \frac{4}{3} \cdot {t}^{2}$

$t = \sqrt{\frac{3 \cdot 7.81}{2}}$

$t = 3.423$ $s$