An object is at rest at #(1 ,2 ,8 )# and constantly accelerates at a rate of #1 m/s^2# as it moves to point B. If point B is at #(3 ,7 ,4 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2017-06-09T02:22:19

It will take 3.66sec.

The distance is #sqrt((1-3)^2+(2-7)^2+(8-4)^2)#=√45
Distance between the two points is √45 m
Solving by equation,
#√45=(1/2)×1×t^2#
#t=3.66#

Answer 2 · 2017-06-09T02:39:55

#t = 3.66# #"s"#

We're asked to find the time #t# it takes the object to travel from two coordinate points with a constant acceleration of #1 "m"/("s"^2)#.

We can use the equation

#Deltax = v_(0x)t + 1/2a_xt^2#

to find this time.

The quantity #Deltax# is the total displacement, which is

#Deltax = sqrt((3"m"-1"m")^2 + (7"m"-2"m")^2 + (4"m"-8"m")^2)#

#=color(red)(6.71# #color(red)("m"#

The intital velocity #v_(0x)# is #0#, because it started from rest.

Plugging in known variables, we have

#color(red)(6.71# #color(red)("m") = (0)t + 1/2(1"m"/("s"^2))t^2#

#t = sqrt((6.71"m")/(1/2"m"/("s"^2))) = color(blue)(3.66# #color(blue)("s"#