An object is at rest at (2 ,1 ,6 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (3 ,4 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Oct 13, 2016

It will take the object 5 seconds to reach point B.

Explanation:

You can use the equation
r = v \Delta t + 1/2 a \Delta t^2
where r is the separation between the two points, v is the initial velocity (here 0, as at rest), a is acceleration and \Delta t is the elapsed time (which is what you want to find).

The distance between the two points is
(3,4,7) - (2,1,6) = (3-2, 4-1, 7-6) = (1,3,1)

r = || (1,3,1) || = \sqrt(1^2 + 3^2 + 1^2) = \sqrt{11} = 3.3166 \text{m}

Substitute r = 3.3166, a = 1/4 and v=0 into the equation given above
3.3166 = 0 + 1/2 1/4 \Delta t^2 Rearrange for \Delta t
\Delta t = \sqrt{(8)(3.3166)}
\Delta t = 5.15 \text{s}

Round to however many decimal places are requested, or to significant figures, of which here there is one, so 5s.