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

Answer:

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#.