An object is at rest at #(7 ,4 ,2 )# and constantly accelerates at a rate of #4/5 m/s# as it moves to point B. If point B is at #(3 ,1 ,9 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Apr 21, 2016

Answer:

Approximately 4.6 seconds

Explanation:

Assuming that the rate of acceleration is #4/5 m/(sec^2)# (that the "squared" was left off the "seconds" of the acceleration.)

Distance between #(7,4,2)# and #(3,1,9)#
#color(white)("XXX")d=sqrt((7-3)^2+(4-1)^2+(2-9)^2)#
#color(white)("XXXX")=sqrt(74)#

The relationship between distance, acceleration (from zero), and time is given by the formula:
#color(white)("XXX")d=1/2a*t^2#

Therefore
#color(white)("XXX")sqrt(74)m=1/(cancel(2)) * cancel(4)^2/5 m/cancel(sec^2) * (t^2 cancel(sec^2))color(white)("/x")#

#color(white)("XXX")t^2=(5sqrt(74))/2#

(using a calculator)
#color(white)("XXX")t=sqrt((5sqrt(74))/2)~~4.637436#