An object is at rest at #(7 ,9 ,4 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(5 ,1 ,8 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Jun 13, 2017

Answer:

#t = 3.24# #"s"#

Explanation:

We're asked to find the time #t# it takes an object to travel a certain distance with a known constant acceleration.

We can use the equation

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

where

  • #Deltax# is the change in position of the object, which is simply the distance between the two coordinate points:

#Deltax = sqrt((5"m" - 7"m")^2 + (1"m" - 9"m")^2 + (8"m" - 4"m")^2)#

#= 9.17# #"m"#

  • #v_(0x)# is the initial velocity, which is #0# since it started from rest,

  • #t# is the time, which is what we're trying to find, and

  • #a_x# is the acceleration, which is #7/4# #"m/s"^2#.

Plugging in known variables we have

#9.17"m" = (0)t + 1/2(7/4"m/s"^2)t^2#

#9.17"m" = (7/8"m/s"^2)t^2#

#t^2 = (9.16"m")/(7/8"m/s"^2)#

#t = sqrt((9.16"m")/(7/8"m/s"^2)) = color(red)(3.24# #color(red)("s"#

The object will travel the distance in #color(red)(3.24# seconds