An object is at rest at #(2 ,9 ,8 )# and constantly accelerates at a rate of #1/5 m/s# as it moves to point B. If point B is at #(6 ,2 ,7 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Jul 4, 2016

Answer:

#t = (66)^(1/4)sqrt(10) s#

Explanation:

I'm assuming the object travels in a straight line. Distance between the two points given by:

#Deltar = sqrt((Deltax)^2+(Deltay)^2+(Deltaz)^2)#

#= sqrt((6-2)^2+(2-9)^2+(7-8)^2) = sqrt(4^2+7^2+1^2) = sqrt(66)#

The object is travelling with constant acceleration along this path, there are no external forces acting on it so equation of motion is:

#(d^2r)/(dt^2) = 1/5#

Integrating gives

#(dr)/(dt) = 1/5t + C_1#

This may be reminiscent of #v = u + at# because, well, it's the same thing. #C_1# denotes initial speed here. As we start from rest, in this case it is equal to zero.

Integrating again gives

#r(t) = 1/10t^2 + C_2#

We know that #r(0) = 0# so # implies C_2 = 0#

#therefore r(t) = 1/10t^2#

The time at which #r = sqrt(66)# is then found by:

#sqrt(66) = 1/10t^2 implies t^2 = 10*sqrt(66)#

Hence #t = (66)^(1/4)sqrt(10) s#

(we discard the negative solution as we are working with times).