An object is at rest at #(1 ,2 ,9 )# 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 23, 2017

Answer:

#3.31"s"#

Explanation:

We're dealing with one-dimensional motion with constant acceleration.

What we can do first is find the displacement #s# during the time interval from point A to point B by using the three-dimensional distance formula:

#s = sqrt((3"m"-1"m")^2 + (1"m"-2"m")^2 + (4"m"-9"m")^2) = 5.48"m"#

Let's look at our known quantities:

Since the object is initially at rest, #v_(0x) = 0#

We'll make the initial position #x_0# be #0#, and the position #x# at time #t# be #5.48"m"#

The acceleration is constant at #1"m"/"s"#

To find the time duration #t# of the displacement #s#, we can use the equation

#x = x_0 + v_(0x)t + 1/2a_xt^2#

And since the initial velocity and position are both #0#, the equation becomes

#x = 1/2a_xt^2#

Here, the variable #x# is analogous to the displacement variable #s#, since they both represent the distance covered. Plugging in the known variables #a_x# and #x#, the time #t# elapsed during the displacement is

#5.48"m" = 1/2(1"m"/("s"^2))t^2#

#t = sqrt((2(5.48"m"/"s"))/(1"m"/("s"^2))) = color(red)(3.31"s"#