# An object is at rest at (7 ,9 ,4 ) and constantly accelerates at a rate of 7/4 m/s^2 as it moves to point B. If point B is at (5 ,1 ,8 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 13, 2017

$t = 3.24$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ it takes an object to travel a certain distance with a known constant acceleration.

We can use the equation

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

where

• $\Delta x$ is the change in position of the object, which is simply the distance between the two coordinate points:

$\Delta x = \sqrt{{\left(5 \text{m" - 7"m")^2 + (1"m" - 9"m")^2 + (8"m" - 4"m}\right)}^{2}}$

$= 9.17$ $\text{m}$

• ${v}_{0 x}$ is the initial velocity, which is $0$ since it started from rest,

• $t$ is the time, which is what we're trying to find, and

• ${a}_{x}$ is the acceleration, which is $\frac{7}{4}$ ${\text{m/s}}^{2}$.

Plugging in known variables we have

9.17"m" = (0)t + 1/2(7/4"m/s"^2)t^2

9.17"m" = (7/8"m/s"^2)t^2

${t}^{2} = \left(9.16 {\text{m")/(7/8"m/s}}^{2}\right)$

t = sqrt((9.16"m")/(7/8"m/s"^2)) = color(red)(3.24 color(red)("s"

The object will travel the distance in color(red)(3.24 seconds