An object is at rest at (1 ,7 ,2 ) and constantly accelerates at a rate of 1 m/s^2 as it moves to point B. If point B is at (3 ,1 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

May 19, 2017

$2.12 s$

Explanation:

We can solve this just a like a one-dimensional constant acceleration problem; the distance covered is

$\Delta x = \sqrt{{3}^{2} + {1}^{2} + {4}^{2}} - \sqrt{{1}^{2} + {7}^{2} + {2}^{2}} = - 2.249 m$

And since the acceleration ${a}_{x} = 1 \frac{m}{{s}^{2}}$, the time $t$ it takes the object to travel $2.249 m$ (in the reverse direction) is

$t = \sqrt{\frac{2 \Delta x}{{a}_{x}}} = \sqrt{\frac{2 \left(2.249 m\right)}{1 \frac{m}{{s}^{2}}}} = 2.12 s$

This is derived from the equation $x = {x}_{0} + {v}_{0 x} t + \frac{1}{2} {a}_{x} {t}^{2}$, where ${v}_{0 x}$ is $0$ (initially at rest), and I made $x$ the distance $2.249 m$ (sign is arbitrary, but needs to be positive to yield a real answer) and ${x}_{0}$ $0$.