An object is at rest at #(2 ,1 ,6 )# and constantly accelerates at a rate of #1/4 m/s^2# as it moves to point B. If point B is at #(3 ,2 ,7 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
May 21, 2017

Answer:

#t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25) = 3.72# #s#

Explanation:

In a 3-dimensional coordinate system, the distance between two points is given by:

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

#r=sqrt((3-2)^2+(2-1)^2+(7-6)^2) =sqrt3 = 1.73# #m#

We can use the equation of motion:

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

Since the initial velocity #u=0#, we can ignore the first term, and rearrange #s= 1/2 at^2# to make #t# the subject.

Note that the term 'r' I was using in the distance calculation above is the same as the distance 's' that the object moves. I don't want the symbol change to be confusing.

#t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25)#

(I used 0.25 rather than 1/4 for ease of writing, but of course it means the same thing)

Therefore #t = 3.72# #s#.