# An object is at rest at (7 ,6 ,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 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 4, 2017

t = 2.138 seconds

#### Explanation:

The distance formula is the square root of the sum of the squares of the changes in x, y, z.
that vector is < 7-5 , 6-5 , 4-7 > = < 2 , 1 , -3 >
The sum of squares is 4 + 1 + 9 = 14
So the distance is ${\left(14\right)}^{\frac{1}{2}}$
Velocity is $\frac{7}{4} \frac{m}{s} ^ 2$
s = v t ==> t = s / v = ${\left(14\right)}^{\frac{1}{2}} / \left(\frac{7}{4}\right)$ = 2.138 sec