An object is at rest at #(1 ,8 ,4 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(1 ,5 ,3 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Sep 15, 2017


2.25 seconds


First thing is to determine the distance to be travelled. We can do this thanks to our old pal Mr. Pythagoras:

#s = sqrt (deltaz^2 + deltay^2 + deltax^2)#

#= sqrt((3-4)^2 + (5-8)^2 + (1-1)^2)#

#= sqrt(10)#

Now, we are given acceleration: #(dv)/dt = 5/4#

Integrate once, and we get v(t):

#v = 5/4t + c1#

constant of integration c1 would represent initial velocity. We are told our mass it at rest at time t = 0, so c1 = 0.

Integrate a second time, and we get position x(t). (x as a function of t).

#x = 5/4 * 1/2 t^2 + c2 = 5/8 t^2 + c2#

We will interpret c2 as the fraction of the distance s (that we calculated above) where the mass will start at. I.e, it's initial position at time t=0. It's at the start point at that time, so it has travelled zero percent of the distance (#sqrt(10)#) as of this time. So we can set c2 to zero.

But now we have everything we need. We know distance, so we can solve for time.

#sqrt(10) = 5/8 t^2#

so, just a little algebra:

#(8sqrt(10))/5 = t^2#

#t = sqrt((8sqrt(10))/5)# = 2.25 seconds (rounding)