An object is at rest at (1 ,8 ,4 ) and constantly accelerates at a rate of 5/4 m/s^2 as it moves to point B. If point B is at (1 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Sep 15, 2017

2.25 seconds

Explanation:

First thing is to determine the distance to be travelled. We can do this thanks to our old pal Mr. Pythagoras:

$s = \sqrt{\delta {z}^{2} + \delta {y}^{2} + \delta {x}^{2}}$

$= \sqrt{{\left(3 - 4\right)}^{2} + {\left(5 - 8\right)}^{2} + {\left(1 - 1\right)}^{2}}$

$= \sqrt{10}$

Now, we are given acceleration: $\frac{\mathrm{dv}}{\mathrm{dt}} = \frac{5}{4}$

Integrate once, and we get v(t):

$v = \frac{5}{4} t + c 1$

constant of integration c1 would represent initial velocity. We are told our mass it at rest at time t = 0, so c1 = 0.

Integrate a second time, and we get position x(t). (x as a function of t).

$x = \frac{5}{4} \cdot \frac{1}{2} {t}^{2} + c 2 = \frac{5}{8} {t}^{2} + c 2$

We will interpret c2 as the fraction of the distance s (that we calculated above) where the mass will start at. I.e, it's initial position at time t=0. It's at the start point at that time, so it has travelled zero percent of the distance ($\sqrt{10}$) as of this time. So we can set c2 to zero.

But now we have everything we need. We know distance, so we can solve for time.

$\sqrt{10} = \frac{5}{8} {t}^{2}$

so, just a little algebra:

$\frac{8 \sqrt{10}}{5} = {t}^{2}$

$t = \sqrt{\frac{8 \sqrt{10}}{5}}$ = 2.25 seconds (rounding)