# An object is at rest at (4 ,5 ,6 ) and constantly accelerates at a rate of 5/3 m/s^2 as it moves to point B. If point B is at (7 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Oct 25, 2017

$\text{ Appro. } 2.26 \sec .$

#### Explanation:

Suppose that, the object is at rest the point (pt.) $A = A \left(4 , 5 , 6\right) .$

Its final position is at the pt. $B = B \left(7 , 5 , 3\right) .$

Hence, using the distance (dist.) formula, the dist. travelled by

the object, i.e.,

$\text{Dist. } A B = \sqrt{{\left(7 - 4\right)}^{2} + {\left(5 - 5\right)}^{2} + {\left(3 - 6\right)}^{2}} , \mathmr{and} ,$

$A B = s = = \sqrt{18.}$

Recall that, $s = u t + \frac{1}{2} a {t}^{2.} \ldots \ldots . \left(\star\right) ,$

where,

$s \text{=dist., "u"=initial velocity, "t"=time, and, "a"=accelaration.}$

In our case, $s = \sqrt{18} , u = 0 , a = \frac{5}{3.}$ Sub.ing in $\left(\star\right) ,$ we get,

$\sqrt{18} = 0 + \frac{1}{2} \cdot \frac{5}{3} {t}^{2} \therefore {t}^{2} = \frac{6 \cdot 3 \sqrt{2}}{5} = \left(3.6\right) \left(1.414\right) ,$

${t}^{2} \approx 5.0912 ,$

$\Rightarrow t \approx \sqrt{5.0912} \approx 2.26 \sec .$