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
II) . The train’s displacement from point A, when it reaches at point C.
II) . The train’s displacement from point A, when it reaches at point B.
II) .The train’s acceleration each times its speed changes.
To plot the graph of the motion of the train, we can use the following information:
- From rest to 20 m/s in 20 seconds: This is a uniform acceleration, so the graph would show a straight line with a positive slope from the origin to the point (20,20).
- At 20 m/s for 60 seconds: Since the train is traveling at a constant speed, the graph would be a horizontal line at y=20.
- From 20 m/s to rest in 40 seconds: This is a uniform deceleration, so the graph would show a straight line with a negative slope from the point (20,20) to the point (60,0).
- At rest for 30 seconds: This would be represented by a horizontal line at y=0 for 30 units of time.
- Reversing direction and accelerating to 10 m/s in 10 seconds: This is a uniform acceleration in the opposite direction, so the graph would show a straight line with a negative slope from the point (90,0) to the point (100,-10).
- At 10 m/s for 30 seconds: This would be represented by a horizontal line at y=-10 for 30 units of time.
- Slowing down to rest in 10 seconds at point B: This is a uniform deceleration, so the graph would show a straight line with a positive slope from the point (130,-10) to the point (140,0).
Here is the plotted graph of the motion of the train:
|
+20 | . B
|
|
|
| . C
0 --------------------
20 60 100 140
A. The train's displacement from point A when it reaches point C can be determined by finding the area under the graph between point A and point C. Using the formula for the area of a trapezoid, we can calculate:
Displacement from A to C = Area of trapezoid = (1/2)(20+20)(60) = 1200 m
B. The train's displacement from point A when it reaches point B can also be determined by finding the area under the graph between point A and point B. Using the formula for the area of a trapezoid, we can calculate:
Displacement from A to B = Area of trapezoid = (1/2)(20+20)(60+30+30) = 2550 m
II. The train's acceleration each time its speed changes can be determined by finding the slope of the graph during those periods of time. From the graph, we can see that the train's acceleration is:
- During the acceleration from rest to 20 m/s in 20 seconds: (20-0)/(20-0) = 1 m/s^2
- During the deceleration from 20 m/s to rest in 40 seconds: (0-20)/(60-20) = -0.5 m/s^2
- During the acceleration from rest to 10 m/s in 10 seconds (reversing direction): (10-0)/(100-90) = 1 m/s^2
- During the deceleration from 10 m/s to rest in 10 seconds: (0-10)/(140-130) = -1 m/s^2