The distance traveled by the student's boxcar that starts from rest and reaches an acceleration and velocity of 13.1 m/s² and 24.5 m/s, respectively, is 22.9 meters.
We can find the distance traveled by the boxcar with the following kinematic equation:
[tex] v_{f}^{2} = v_{i}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 24.5 m/s
[tex] v_{i}[/tex]: is the initial velocity = 0 (it starts from rest)
a: is the acceleration = 13.1 m/s²
d: is the distance =?
By solving the above equation for d, we have:
[tex] d = \frac{v_{f}^{2} - v_{i}^{2}}{2a} = \frac{(24.5 m/s)^{2}}{2*13.1 m/s^{2}} = 22.9 m [/tex]
Therefore, the boxcar will travel 22.9 meters.
You can find more about distance and acceleration here: https://brainly.com/question/14363745?referrer=searchResults
I hope it helps you!
