A train travelling along a straight truck start from rest at point A and accelerated uniformly to 20 m/s in 20 second. It travels at this speed for 60 second. Then slow down uniformly to rest in 40 second at a point C. it stays at rest point C for 30 seconds. Then reverses direction accelerating uniformly to 10 m/s in 10 second. It travels at this speed for 30 second. Then slows down uniformly to rest in 10 second, when it reaches point B.
a) Plot a graph of the motion of the train.
b) Use your graph to accelerate
1) The train’s displacement from point A, when it reaches at point C.
2) The train’s displacement from point A, when it reaches at point B.
3) The train’s acceleration each times its speed changes.
To plot the motion of the train, we can use a graph with time on the x-axis and velocity on the y-axis.
a) Plotting the motion of the train:
First, let's divide the motion into four parts:
1) Accelerating from rest to 20 m/s in 20 seconds.
2) Traveling at a constant speed of 20 m/s for 60 seconds.
3) Decelerating from 20 m/s to rest in 40 seconds.
4) Reversing direction, accelerating from rest to 10 m/s in 10 seconds, traveling at a constant speed of 10 m/s for 30 seconds, and then decelerating to rest in 10 seconds.
Here is the plotted graph:
```
|
20 |________________
| |
10 | | ____
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
0 |________________|__|__|_____________________
0 20 40 60 80 90 100 110 120 130 140 150
(s)
```
b) Using the graph to calculate:
1) The train's displacement from point A when it reaches point C:
From the graph, we can see that the train travels at a constant speed of 20 m/s for 60 seconds. The area under the curve for this section represents the displacement. The area is a rectangle with a base of 60 seconds and a height of 20 m/s.
Displacement = base * height = 60s * 20m/s = 1200 meters
2) The train's displacement from point A when it reaches point B:
Similarly, the train travels at a constant speed of 10 m/s for 30 seconds. The area under the curve for this section is a rectangle with a base of 30 seconds and a height of 10 m/s.
Displacement = base * height = 30s * 10m/s = 300 meters
3) The train's acceleration each time its speed changes:
- Acceleration when the train is accelerating from rest to 20 m/s in 20 seconds:
Acceleration = (change in velocity) / (time) = (20m/s - 0m/s) / 20s = 1 m/s²
- Acceleration when the train is decelerating from 20 m/s to rest in 40 seconds:
Acceleration = (change in velocity) / (time) = (0m/s - 20m/s) / 40s = -0.5 m/s²
- Acceleration when the train is reversing, accelerating from rest to 10 m/s in 10 seconds:
Acceleration = (change in velocity) / (time) = (10m/s - 0m/s) / 10s = 1 m/s²
- Acceleration when the train is decelerating from 10 m/s to rest in 10 seconds:
Acceleration = (change in velocity) / (time) = (0m/s - 10m/s) / 10s = -1 m/s²