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 m/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 6.0m/s . Once free of this area, it speeds up to 14m/s in 8.0 seconds. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 seconds to reach a higher cruising speed.
what is the final speed?
To find the final speed of the train, we need to calculate the total distance traveled and then use the equation of motion to determine the final speed.
First, let's calculate the distance traveled during each acceleration phase.
In the congested part of town:
Initial speed, u1 = 6.0 m/s
Acceleration, a = 1.3 m/s^2
Time, t1 = ? (unknown)
Using the equation of motion: v = u + at, where v is the final velocity,
We can rearrange the equation to solve for time:
t1 = (v - u1) / a
t1 = (0 - 6.0) / 1.3
t1 = -6.0 / 1.3 (Note: In this context, time cannot be negative, so it means the train is decelerating or slowing down)
t1 ≈ -4.62 seconds
The train decelerates for about 4.62 seconds in the congested part of town.
Next, let's calculate the distance traveled during the second acceleration phase.
Initial speed, u2 = 6.0 m/s (since the train was moving at this speed after the congested part)
Acceleration, a = 1.3 m/s^2
Time, t2 = 16 seconds
Using the equation of motion: s = u2t2 + 0.5at2^2, where s is the distance traveled,
s2 = (6.0)(16) + 0.5(1.3)(16)^2
s2 = 96 + 0.5(1.3)(256)
s2 = 96 + 166.4
s2 = 262.4 meters
The train travels approximately 262.4 meters during the second acceleration phase.
Now, let's calculate the distance traveled during the constant-speed phase.
Initial speed for the constant-speed phase = 14.0 m/s (since this is the speed after the second acceleration phase)
Time during the constant-speed phase = 16.0 seconds (The question doesn't mention any acceleration, so we assume the train maintains a constant speed for this duration)
Using the equation of motion: s = ut,
s3 = (14.0)(16.0)
s3 = 224.0 meters
The train travels 224.0 meters during the constant-speed phase.
Lastly, to find the total distance traveled, we add the distances traveled during each phase.
Total distance traveled = s1 + s2 + s3
Total distance = 0 + 262.4 + 224.0
Total distance ≈ 486.4 meters
Finally, we can calculate the final speed using the equation of motion: v = u + at.
Initial speed for the final acceleration phase = 14.0 m/s (since this is the speed after the constant-speed phase)
Acceleration, a = 1.3 m/s^2
Time taken for the final acceleration phase = 16.0 seconds (as mentioned in the question)
Using the equation of motion:
v^2 = u^2 + 2as
v^2 = (14.0)^2 + 2(1.3)(486.4)
v^2 = 196 + 1262.24
v^2 ≈ 1458.24
v ≈ √1458.24
v ≈ 38.2 m/s
Therefore, the final speed of the train is approximately 38.2 m/s.