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.
iii) The train’s acceleration each times its speed changes.
To plot the graph of the train's motion, we will use the following variables:
- t: time in seconds
- v: velocity in m/s
- a: acceleration in m/s²
- s: displacement in meters from point A
Using this information, we can break down the train's motion into four intervals:
Interval 1 (0 ≤ t ≤ 20 seconds):
- The train starts from rest, so v = 0 m/s initially.
- The train accelerates uniformly, so a is constant.
- Given that v = 20 m/s at t = 20 seconds, we can find the acceleration:
v = u + at
20 = 0 + a * 20
a = 1 m/s² (acceleration during the first interval)
Interval 2 (20 ≤ t ≤ 80 seconds):
- The train maintains a constant speed of 20 m/s, so v = 20 m/s throughout this interval.
- Since the velocity is constant, the acceleration is 0 m/s².
Interval 3 (80 ≤ t ≤ 120 seconds):
- The train slows down uniformly, so a is constant.
- Given that the velocity reduces to 0 m/s at t = 120 seconds and the deceleration interval is 40 seconds, we can find the acceleration:
v = u + at
0 = 20 + a * 40
a = -0.5 m/s² (deceleration during the third interval)
Interval 4 (120 ≤ t ≤ 160 seconds):
- The train stays at rest, so v = 0 m/s throughout this interval.
- Since the velocity is constant, the acceleration is 0 m/s².
Now, let's plot the graph for the train's motion:
Interval 1: (0 ≤ t ≤ 20 seconds)
- The area under the graph represents the displacement (s) from point A.
- The area of a trapezoid can be calculated using the formula: Area = 1/2 * (a1 + a2) * h, where a1 and a2 are the lengths of the parallel sides and h is the height.
- The trapezoid has a base of 20 seconds and heights of 0 and 20 m/s, respectively.
- The displacement (s) in this interval can be calculated as:
s1 = 1/2 * 0.5 * 20 * 20 = 200 meters
Interval 2: (20 ≤ t ≤ 80 seconds)
- The velocity remains constant at 20 m/s, so the displacement is simply the product of velocity and time:
s2 = 20 * (80 - 20) = 1200 meters
Interval 3: (80 ≤ t ≤ 120 seconds)
- The area under the graph represents the displacement (s) from point A.
- The trapezoid has a base of 40 seconds and heights of 20 and 0 m/s, respectively.
- The displacement (s) in this interval can be calculated as:
s3 = 1/2 * (20 + 0) * 40 = 400 meters
Interval 4: (120 ≤ t ≤ 160 seconds)
- The velocity is 0 m/s, so the displacement is 0 meters.
Now let's answer the questions:
A) Graph of motion: The graph will consist of a straight line for the first and third intervals (acceleration and deceleration) and a horizontal line for the second and fourth intervals (constant velocity and rest).
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|
200 | /
|/
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/|
400 |
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/|
|
120 160
time (s)
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B) Accelerations at each speed change:
i) The train's displacement from point A when it reaches point C:
- At point C (t = 120 seconds), the train has a displacement of s1 + s2 + s3 = 200 + 1200 + 400 = 1800 meters.
ii) The train's displacement from point A when it reaches point B:
- At point B (t = 160 seconds), the train's displacement is the same as in point C, so it is still 1800 meters.
iii) The train's acceleration each time its speed changes:
- Acceleration during the first interval: a = 1 m/s²
- Acceleration during the third interval: a = -0.5 m/s²