Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magnitude of the maximum acceleration is typically 1.3 {\rm m}/{\rm s}^{2}, but the driver will usually maintain a constant acceleration that is less than the maximum. A train travels through a congested part of town at 4.0m/s . Once free of this area, it speeds up to 11m/s in 8.0 {\rm s}. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 {\rm s} to reach a higher cruising speed.
What is the questi0n?
What is the final speed?
38
wrong
To solve this problem, we need to find the acceleration of the train when it speeds up to reach 11 m/s in 8.0 s, as well as the acceleration when it reaches its cruising speed.
Let's start with the first part of the problem. The train initially travels at 4.0 m/s and speeds up to 11 m/s in 8.0 s.
We can use the following kinematic equation to find the acceleration:
vf = vi + at
where:
vf is the final velocity (11 m/s),
vi is the initial velocity (4.0 m/s),
a is the acceleration, and
t is the time (8.0 s).
Rearranging the equation to solve for the acceleration (a), we get:
a = (vf - vi) / t
Substituting the given values, we have:
a = (11 m/s - 4.0 m/s) / 8.0 s
Simplifying the equation, we get:
a = 7.0 m/s / 8.0 s
a = 0.875 m/s^2
So, the acceleration of the train when it speeds up to reach 11 m/s in 8.0 s is 0.875 m/s^2.
Now let's move on to the second part of the problem. The train accelerates for another 16 s to reach its cruising speed.
Given that the magnitude of the maximum acceleration is typically 1.3 m/s^2, we can assume that the acceleration during this phase is also 1.3 m/s^2.
Therefore, the acceleration of the train when it reaches its cruising speed is 1.3 m/s^2.
To summarize:
- The acceleration of the train when it speeds up to reach 11 m/s in 8.0 s is 0.875 m/s^2.
- The acceleration of the train when it reaches its cruising speed is 1.3 m/s^2.