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
i) 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 a velocity-time graph.
First, let's divide the timeline into different phases and calculate the velocities at various points.
Phase 1: Acceleration from rest to 20 m/s in 20 seconds
- Velocity at the end of this phase = 20 m/s
- The slope of the velocity-time graph during this phase is constant and equal to the acceleration. Since the train starts from rest and reaches 20 m/s in 20 seconds, the acceleration is 20 m/s / 20 s = 1 m/s^2.
- During this phase, the train covers a distance equal to the area under the velocity-time graph, which is also the displacement. The displacement can be calculated using the formula: displacement = (initial velocity + final velocity) / 2 * time = (0 + 20) / 2 * 20 = 200 m.
Phase 2: Constant velocity of 20 m/s for 60 seconds
- The velocity-time graph will be a straight line at 20 m/s for 60 seconds.
Phase 3: Deceleration from 20 m/s to rest in 40 seconds
- Velocity at the end of this phase = 0 m/s
- The slope of the velocity-time graph during this phase is constant and equal to the deceleration. Since the train starts from 20 m/s and reaches rest in 40 seconds, the deceleration is -20 m/s / 40 s = -0.5 m/s^2 (negative sign indicates deceleration).
- Similar to phase 1, the displacement during this phase can be calculated as (initial velocity + final velocity) / 2 * time = (20 + 0) / 2 * 40 = 400 m.
Phase 4: Rest for 30 seconds
- The velocity-time graph will be a horizontal line at 0 m/s for 30 seconds.
Phase 5: Reversal and acceleration from rest to 10 m/s in 10 seconds
- Velocity at the end of this phase = 10 m/s
- The slope of the velocity-time graph during this phase is constant and equal to the acceleration. Since the train starts from rest and reaches 10 m/s in 10 seconds, the acceleration is 10 m/s / 10 s = 1 m/s^2 (notice that acceleration is positive now).
- Similar to phase 1, the displacement during this phase can be calculated as (initial velocity + final velocity) / 2 * time = (0 + 10) / 2 * 10 = 50 m.
Phase 6: Constant velocity of 10 m/s for 30 seconds
- The velocity-time graph will be a straight line at 10 m/s for 30 seconds.
Phase 7: Deceleration from 10 m/s to rest in 10 seconds
- Velocity at the end of this phase = 0 m/s
- The slope of the velocity-time graph during this phase is constant and equal to the deceleration. Since the train starts from 10 m/s and reaches rest in 10 seconds, the deceleration is -10 m/s / 10 s = -1 m/s^2 (negative sign indicates deceleration).
- Similar to phase 1, the displacement during this phase can be calculated as (initial velocity + final velocity) / 2 * time = (10 + 0) / 2 * 10 = 50 m.
Now, let's plot the velocity-time graph with suitable scales:
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20 _____| ___________________
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10 __| |_____|_______
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|_____|_____|_______
0 20 40 60
From the graph, we can see that the train's displacement from point A when it reaches point C is 600 m (200 m + 400 m).
Similarly, the train's displacement from point A when it reaches point B can be calculated as the total displacement up to point B, which is equal to the area under the entire graph. This area can be divided into two triangles and a rectangle.
Displacement from A to C = 200 m + 400 m = 600 m
Displacement from C to B = 0 m (since the train stays at rest at point C)
Displacement from B to A = 50 m (area of the first triangle)
Total displacement from A to B = 600 m + 0 m + 50 m = 650 m
Next, let's calculate the train's acceleration each time its speed changes.
Acceleration during phase 1 (from rest to 20 m/s) = 1 m/s^2
Acceleration during phase 3 (from 20 m/s to rest) = -0.5 m/s^2
Acceleration during phase 5 (from rest to 10 m/s) = 1 m/s^2
Acceleration during phase 7 (from 10 m/s to rest) = -1 m/s^2
To summarize:
A) The graph of the motion of the train is a piecewise velocity-time graph with different phases representing acceleration, constant velocity, and deceleration.
B) Accelerations:
- Phase 1: 1 m/s^2
- Phase 3: -0.5 m/s^2
- Phase 5: 1 m/s^2
- Phase 7: -1 m/s^2
i) The train's displacement from point A when it reaches point C is 600 m.
ii) The train's displacement from point A when it reaches point B is 650 m.