Question

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/s2 , 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 13m/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?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The final speed of the train is 46 m/s.

Step-by-step explanation:

The final speed of the train can be found by calculating the total distance traveled during each acceleration phase and dividing it by the total time taken. In the first phase, the train travels through a congested part of town at 4.0 m/s and maintains this speed, so there is no acceleration. In the second phase, the train speeds up to 13 m/s in 8.0 s, so the average acceleration can be calculated using the formula: average acceleration = change in velocity / time = (final velocity - initial velocity) / time = (13 m/s - 4.0 m/s) / 8.0 s = 9.0 m/s^2.

Using this average acceleration, we can find the distance covered in the second phase using the formula: distance = initial velocity * time + 0.5 * average acceleration * time^2 = 4.0 m/s * 8.0 s + 0.5 * 9.0 m/s^2 * (8.0 s)^2 = 32 m + 288 m = 320 m.

In the third phase, the train accelerates for another 16 s with the same acceleration of 9.0 m/s^2. Using the formula for distance covered during acceleration, we can calculate: distance = initial velocity * time + 0.5 * average acceleration * time^2 = 13 m/s * 16 s + 0.5 * 9.0 m/s^2 * (16 s)^2 = 208 m + 576 m = 784 m.

Adding together the distances traveled in each phase, the total distance covered by the train is 320 m + 784 m = 1104 m. To find the final speed, we can divide the total distance by the total time: final speed = total distance / total time = 1104 m / (8.0 s + 16 s) = 1104 m / 24 s = 46 m/s.

Answer 2

During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s