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.

Question

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.

Lily Adams · Accepted Answer

#3.31"s"#

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"#

Layla Ahmed · Answer

undefined To find the time it takes for the object to reach point B, use the formula:

$$ \text{time} = \sqrt{\frac{2 \cdot \text{distance}}{\text{acceleration}}} $$

First, calculate the distance between the initial point and point B using the distance formula in three dimensions:

$$ \text{distance} = \sqrt{(3 - 1)^2 + (1 - 2)^2 + (4 - 9)^2} $$

$$ \text{distance} = \sqrt{2^2 + (-1)^2 + (-5)^2} $$

$$ \text{distance} = \sqrt{4 + 1 + 25} $$

$$ \text{distance} = \sqrt{30} $$

Now, plug the values into the time formula:

$$ \text{time} = \sqrt{\frac{2 \cdot \sqrt{30}}{1}} $$

$$ \text{time} = \sqrt{2 \cdot \sqrt{30}} $$

$$ \text{time} \approx \sqrt{2} \times \sqrt{\sqrt{30}} $$

$$ \text{time} \approx \sqrt[4]{60} $$

$$ \text{time} \approx 2.89 \, \text{s} $$