An object is at rest at #(3 ,5 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(9 ,9 ,8 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Jun 3, 2017

Answer:

#3.9# #"s"#

Explanation:

We're asked to find the time it takes for the object to move from #(3, 5, 1)# to #(9, 9, 8)# at a constant acceleration of #4/3"m"/("s"^2)#. (This is one-dimensional motion, as it's moving in a straight line, regardless of the fact that the coordinates are three-dimensional.)

To do this, we can use the equation

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

Our known quantities are

  • #Deltax# is the change in position of the object from the two coordinate points. We can find the distance between these two coordinates and call this the change in position:

#Deltax = sqrt((9"m"-3"m")^2 + (9"m"-5"m")^2 + (8"m"-1"m")^2)#

#= color(red)(10.0"m"#

  • #v_(0x)# is the initial velocity of the object. Since it's originally at rest, this value is zero.

  • #a_x# is the object's (constant) acceleration, which is #4/3"m"/("s"^2)#

Plugging in our known values, we have

#color(red)(10.0"m") = (0)t + 1/2(4/3"m"/("s"^2))t^2#

#color(red)(10.0"m") = (2/3"m"/("s"^2))t^2#

#t = sqrt((10.0"m")/(2/3"m"/("s"^2))) = color(blue)(3.9# #color(blue)("s"#

The object will travel the distance from #(3, 5, 1)# to #(9, 9, 8)# in #3.9# seconds.