An object is at rest at #(8 ,3 ,3 )# and constantly accelerates at a rate of #7/4 ms^-2# as it moves to point B. If point B is at #(6 ,3 ,4 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Jul 8, 2017

First step is to find the distance between the two points: #r~=2.24# #m#

Second step is to find the time taken: #t=1.6# #s#

Explanation:

To find the distance between two points in 3D space we have:

#r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

#=sqrt((6-8)^2+(3-3)^2+(4-3)^2)=sqrt((-2)^2+(0)^2+(1)^2)#

Therefore:

#r=sqrt5~=2.24#

Now to find the time taken, we can use:

#d=ut+1/2at^2#

Since the object is initially at rest, the #ut# term goes to #0#.

Rearranging:

#t=sqrt((2d)/a)#

Note that I have used both #d# and #r# because of the way I remember formulae, but we're talking about the same distance - the distance between the points.

#t=sqrt((2xx2.24)/(7/4))= 1.6# #s#