# An object is at rest at (1 ,2 ,8 ) and constantly accelerates at a rate of 1 m/s^2 as it moves to point B. If point B is at (3 ,7 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 9, 2017

It will take 3.66sec.

#### Explanation:

The distance is $\sqrt{{\left(1 - 3\right)}^{2} + {\left(2 - 7\right)}^{2} + {\left(8 - 4\right)}^{2}}$=√45
Distance between the two points is √45 m
Solving by equation,
√45=(1/2)×1×t^2
$t = 3.66$

Jun 9, 2017

$t = 3.66$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ it takes the object to travel from two coordinate points with a constant acceleration of 1 "m"/("s"^2).

We can use the equation

$\Delta x = {v}_{0 x} t + \frac{1}{2} {a}_{x} {t}^{2}$

to find this time.

The quantity $\Delta x$ is the total displacement, which is

$\Delta x = \sqrt{{\left(3 \text{m"-1"m")^2 + (7"m"-2"m")^2 + (4"m"-8"m}\right)}^{2}}$

=color(red)(6.71 color(red)("m"

The intital velocity ${v}_{0 x}$ is $0$, because it started from rest.

Plugging in known variables, we have

color(red)(6.71 color(red)("m") = (0)t + 1/2(1"m"/("s"^2))t^2

t = sqrt((6.71"m")/(1/2"m"/("s"^2))) = color(blue)(3.66 color(blue)("s"