# An object is at rest at (7 ,4 ,2 ) and constantly accelerates at a rate of 4/5 m/s as it moves to point B. If point B is at (3 ,1 ,9 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Apr 21, 2016

Approximately 4.6 seconds

#### Explanation:

Assuming that the rate of acceleration is $\frac{4}{5} \frac{m}{{\sec}^{2}}$ (that the "squared" was left off the "seconds" of the acceleration.)

Distance between $\left(7 , 4 , 2\right)$ and $\left(3 , 1 , 9\right)$
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{\left(7 - 3\right)}^{2} + {\left(4 - 1\right)}^{2} + {\left(2 - 9\right)}^{2}}$
$\textcolor{w h i t e}{\text{XXXX}} = \sqrt{74}$

The relationship between distance, acceleration (from zero), and time is given by the formula:
$\textcolor{w h i t e}{\text{XXX}} d = \frac{1}{2} a \cdot {t}^{2}$

Therefore
$\textcolor{w h i t e}{\text{XXX")sqrt(74)m=1/(cancel(2)) * cancel(4)^2/5 m/cancel(sec^2) * (t^2 cancel(sec^2))color(white)("/x}}$

$\textcolor{w h i t e}{\text{XXX}} {t}^{2} = \frac{5 \sqrt{74}}{2}$

(using a calculator)
$\textcolor{w h i t e}{\text{XXX}} t = \sqrt{\frac{5 \sqrt{74}}{2}} \approx 4.637436$