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.

1 Answer
Jan 11, 2018

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#.

Explanation:

First step is to find the distance between the two points:

#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#

Now we know:

#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#

We have the equation of motion:

#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#