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

1 Answer
May 19, 2017

Answer:

#2.12s#

Explanation:

We can solve this just a like a one-dimensional constant acceleration problem; the distance covered is

#Deltax = sqrt (3^2 + 1^2 + 4^2) - sqrt(1^2 + 7^2 + 2^2)= -2.249 m#

And since the acceleration #a_x = 1m/(s^2)#, the time #t# it takes the object to travel #2.249m# (in the reverse direction) is

#t = sqrt((2Deltax)/(a_x)) = sqrt((2(2.249m))/(1m/(s^2))) = 2.12 s#

This is derived from the equation #x = x_0 + v_(0x)t + 1/2a_xt^2#, where #v_(0x)# is #0# (initially at rest), and I made #x# the distance #2.249m# (sign is arbitrary, but needs to be positive to yield a real answer) and #x_0# #0#.