An object is at rest at $(5, 1, 1)$ $(5 ,1 ,1 )$ and constantly accelerates at a rate of $\frac{2}{5} \frac{m}{s}$ $2/5 m/s$ as it moves to point B. If point B is at $(6, 9, 7)$ $(6 ,9 ,7 )$ , 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 · 2016-11-05T16:17:29

$t \approx 7 s$ $t ~~ 7 s$

I am going to assume that there is a small typographical error in the units of the acceleration and it is $\frac{2}{5} \frac{m}{s^{2}}$ $2/5m/s^2$

The distance, d, from Point $A = (5, 1, 1)$ $A = (5,1,1)$ to point $B = (6, 9, 7)$ $B = (6,9,7)$ is

$d = \sqrt{{(6 - 5)}^{2} + {(9 - 1)}^{2} + {(7 - 1)}^{2}}$ $d =sqrt((6 -5)^2 + (9 - 1)^2 + (7 - 1)^2)$

$d = \sqrt{{(1)}^{2} + {(8)}^{2} + {(6)}^{2}}$ $d =sqrt((1)^2 + (8)^2 + (6)^2)$

$d = \sqrt{1 + 64 + 36}$ $d =sqrt(1 + 64 + 36)$

$d = \sqrt{101} m$ $d =sqrt(101) m$

The equation for distance traveled under constant acceleration is:

$d = (\frac{1}{2}) a t^{2}$ $d = (1/2)at^2$

Substitute $\sqrt{101} m$ $sqrt(101) m$ for d and $\frac{2}{5} \frac{m}{s^{2}}$ $2/5m/s^2$ for a, and then solve for t:

$\sqrt{101} m = (\frac{1}{2}) (\frac{2}{5} \frac{m}{s^{2}}) t^{2}$ $sqrt(101) m = (1/2)(2/5m/s^2)t^2$

$5 \sqrt{101} s^{2} = t^{2}$ $5sqrt(101)s^2 = t^2$

$t \approx \pm 7 s$ $t ~~ +-7 s$

But negative time does not make sense, therefore, make it only positive:

$t \approx 7 s$ $t ~~ 7 s$

An object is at rest at (5,1,1)(5 ,1 ,1 ) and constantly accelerates at a rate of 25ms2/5 m/s as it moves to point B. If point B is at (6,9,7)(6 ,9 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.