An object is at rest at #(1 ,2 ,1 )# and constantly accelerates at a rate of #1/3# #ms^-2# as it moves to point B. If point B is at #(4 ,4 ,5 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

An object is at rest at #(1 ,2  ,1  )# and constantly accelerates at a rate of #1/3# #ms^-2# as it moves to point B.  If point B is at #(4   ,4   ,5   )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Olivia Askew · Accepted Answer

The distance between the two points is #5.4# #m# and the time taken to move from one to the other is #5.7# #s#.

Finding the distance between the two points is the first step. #s=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)# #=sqrt((4-1)^2+(4-2)^2+(5-1)^2)=sqrt(3^2+2^2+4^2)# #=sqrt(9+4+16)=sqrt(29)=5.4# #m# We now understand: #u=0# #ms^-1# (since the object was initially at rest) #a=1/3# #ms^-2# #s=5.4# #m# (some people use #d# for distance) #t=?# #s# The equation of motion is as follows: #s=ut+1/2at^2# The #ut# term goes to 0 because #u=0#, which makes life simpler: #s=1/2at^2# Rearranging to make #t# the subject: #t=sqrt((2s)/a)=sqrt((2xx5.4)/(1/3))=sqrt 32.4=5.7# #s#

Carson Alexander · Answer

undefined Using the kinematic equation $ s = ut + \frac{1}{2}at^2 $, where $ s $ is the displacement, $ u $ is the initial velocity (which is 0 since the object is at rest), $ a $ is the acceleration, and $ t $ is the time:

For each coordinate (x, y, z):
- $ s_x = x_B - x_A $
- $ s_y = y_B - y_A $
- $ s_z = z_B - z_A $

Substitute these values into the kinematic equation, and solve for time ($ t $) separately for each coordinate. The resulting time for all coordinates will be the same.