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.
Plot a graph of the motion of the train.
Use your graph to accelerate
The train’s displacement from point A, when it reaches at point C.
The train’s displacement from point A, when it reaches at point B.
The train’s acceleration each times its speed changes.
To plot the graph of the motion of the train, we can use a graph with time on the x-axis and velocity on the y-axis.
First, let's divide the given time intervals into smaller intervals to plot the graph accurately. We will take intervals of 1 second for this example.
1. From 0 to 20 seconds (acceleration):
Time: 0, 1, 2, ..., 19, 20
Velocity: 0, 1, 2, ..., 19, 20
2. From 20 to 80 seconds (constant speed):
Time: 21, 22, ..., 79, 80
Velocity: 20, 20, ..., 20, 20
3. From 80 to 120 seconds (deceleration):
Time: 81, 82, ..., 119, 120
Velocity: 20, 19, ..., 1, 0
4. From 120 to 150 seconds (rest at point C):
Time: 121, 122, ..., 149, 150
Velocity: 0, 0, ..., 0, 0
5. From 150 to 160 seconds (reversing direction, acceleration):
Time: 151, 152, ..., 159, 160
Velocity: 0, 1, ..., 9, 10
6. From 160 to 190 seconds (constant speed in reverse):
Time: 161, 162, ..., 189, 190
Velocity: 10, 10, ..., 10, 10
7. From 190 to 200 seconds (deceleration to rest at point B):
Time: 191, 192, ..., 199, 200
Velocity: 10, 9, ..., 1, 0
Now let's calculate the displacement from point A when the train reaches point C and point B.
1. Displacement from A to C:
Displacement during acceleration phase = 0.5 * 20 * 20 = 200 m (using the formula displacement = 0.5 * acceleration * time^2)
Displacement during constant speed phase = 20 * 60 = 1200 m (since the velocity is constant)
Displacement during deceleration phase = 0.5 * 0 * 40 = 0 m (since the final velocity is 0)
Total displacement from A to C = 200 + 1200 + 0 = 1400 m
2. Displacement from A to B:
Displacement during acceleration phase (reversing direction) = 0.5 * 10 * 10 = 50 m
Displacement during constant speed phase (in reverse) = 10 * 30 = 300 m
Displacement during deceleration phase = 0.5 * 0 * 10 = 0 m
Total displacement from A to B = Displacement from A to C + Displacement during acceleration phase (reversing direction) + Displacement during constant speed phase (in reverse) + Displacement during deceleration phase = 1400 + 50 + 300 + 0 = 1750 m
Now let's calculate the acceleration each time the train's speed changes.
Acceleration during acceleration phase from rest to 20 m/s:
Acceleration = Change in velocity / Time taken = (20 - 0) / 20 = 1 m/s^2
Acceleration during deceleration phase from 20 m/s to rest:
Acceleration = Change in velocity / Time taken = (0 - 20) / 40 = -0.5 m/s^2 (negative since it's deceleration)
Acceleration during acceleration phase (reversing direction) from rest to 10 m/s:
Acceleration = Change in velocity / Time taken = (10 - 0) / 10 = 1 m/s^2
Acceleration during deceleration phase from 10 m/s (in reverse) to rest:
Acceleration = Change in velocity / Time taken = (0 - 10) / 10 = -1 m/s^2 (negative since it's deceleration)
Now we have plotted the graph of the motion of the train, calculated the displacement from point A to point C and point B, and also determined the acceleration each time the train's speed changes.