An object is at rest at (2 ,9 ,8 ) and constantly accelerates at a rate of 1/5 m/s as it moves to point B. If point B is at (6 ,2 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 4, 2016

$t = {\left(66\right)}^{\frac{1}{4}} \sqrt{10} s$

Explanation:

I'm assuming the object travels in a straight line. Distance between the two points given by:

$\Delta r = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} + {\left(\Delta z\right)}^{2}}$

$= \sqrt{{\left(6 - 2\right)}^{2} + {\left(2 - 9\right)}^{2} + {\left(7 - 8\right)}^{2}} = \sqrt{{4}^{2} + {7}^{2} + {1}^{2}} = \sqrt{66}$

The object is travelling with constant acceleration along this path, there are no external forces acting on it so equation of motion is:

$\frac{{d}^{2} r}{{\mathrm{dt}}^{2}} = \frac{1}{5}$

Integrating gives

$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{5} t + {C}_{1}$

This may be reminiscent of $v = u + a t$ because, well, it's the same thing. ${C}_{1}$ denotes initial speed here. As we start from rest, in this case it is equal to zero.

Integrating again gives

$r \left(t\right) = \frac{1}{10} {t}^{2} + {C}_{2}$

We know that $r \left(0\right) = 0$ so $\implies {C}_{2} = 0$

$\therefore r \left(t\right) = \frac{1}{10} {t}^{2}$

The time at which $r = \sqrt{66}$ is then found by:

$\sqrt{66} = \frac{1}{10} {t}^{2} \implies {t}^{2} = 10 \cdot \sqrt{66}$

Hence $t = {\left(66\right)}^{\frac{1}{4}} \sqrt{10} s$

(we discard the negative solution as we are working with times).