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

1 Answer
David G.
May 21, 2016

First step is to find the distance between the points, which is approximately 9.2 m. Then the time taken can be calculated, as shown below, to be 6.8 s.

Explanation:

Distance between the points:

r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)
=sqrt((6-2)^2+(9-1)^2+(7-1)^2)=sqrt(4^2+8^2+6^2)
=sqrt(16+64+4)=sqrt(84)~~9.2 m

Finding the time taken, given that the object is at rest (u=0):

d=ut+1/2at^2=1/2at^2 (when u=0)

Rearranging:

t=sqrt((2d)/a)=sqrt((2*9.2)/(2/5))=sqrt46~~6.8 s

Related questions
See all questions in Acceleration
Impact of this question
1450 views around the world
You can reuse this answer
Creative Commons License