Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 5.0 m/s. Once free of this area, it speeds up to 13 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed.
What is the final speed?
no its 25
Well, the train's journey sounds like quite the roller coaster ride! Let's break it down. We have the initial speed of 5.0 m/s, then it speeds up to 13 m/s in 8.0 seconds, and finally accelerates for another 16 seconds to reach the cruising speed.
Since the acceleration is constant throughout, we can use the equation:
final speed = initial speed + (acceleration × time)
First, let's find the acceleration during the second part of the journey. We know that the initial speed is 13 m/s, the final speed is what we're looking for, and the time is 16 seconds.
So, final speed = 13 m/s + (acceleration × 16 s)
Now, to get the final speed, I need to know the acceleration. Mind sharing that with me?
To determine the final speed of the train, we first need to calculate its acceleration during the speed up periods.
Given:
Initial speed (v0) = 5.0 m/s
Final speed (v) = 13 m/s
Time taken to reach final speed (t) = 8.0 s
We can use the equation:
v = v0 + at
Rearranging the equation to solve for acceleration (a):
a = (v - v0) / t
Substituting the given values:
a = (13 m/s - 5.0 m/s) / 8.0 s
Calculating this:
a = 8.0 m/s / 8.0 s
a = 1.0 m/s²
Now, we know that the train accelerates with the same acceleration for the next 16 seconds. So, we can calculate the speed gained during this period:
t = 16 s
v0 = 13 m/s
a = 1.0 m/s²
Using the same equation as before:
v = v0 + at
v = 13 m/s + 1.0 m/s² * 16 s
Calculating this:
v = 13 m/s + 16 m/s
v = 29 m/s
Therefore, the final speed of the train after the second acceleration is 29 m/s.
a = (13 - 5) / 8 = 1 m/s^2
v = 13 + a t = 13 + 1 * 16
= 29 m/s