A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 14.0s. It runs at a constant speed for 70.0 s and then slows down at a rate of -3.50 m/s2 until it stops at the next station.

a) Sketch a position vs. time graph for this problem.
b) Sketch a velocity vs. time graph for this problem.
c) Find the total distance covered by the train.

a)

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Position |_______________________
Station 1 Station 2

b)

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Velocity| \ __
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Station 1 Station 2

c) To find the total distance covered by the train, we need to calculate the distances during each phase of motion.

During acceleration:
Using the equation for displacement during constant acceleration:
d = v₀t + (1/2)at²

Initial velocity (v₀) = 0
Acceleration (a) = 1.60 m/s²
Time (t) = 14.0 s

d₁ = (0)(14.0) + (1/2)(1.60)(14.0)²
d₁ = (0) + (0.5)(1.60)(196)
d₁ = 0 + 313.6
d₁ = 313.6 m

During constant speed:
The train runs at a constant speed for 70.0 s, so its velocity remains the same.
Velocity (v) = (1.60 m/s²)(14.0 s)
v = 22.4 m/s

d₂ = v × t
d₂ = 22.4 × 70.0
d₂ = 1,568 m

During deceleration:
Using the same equation as acceleration, but with a negative acceleration value:
Acceleration (a) = -3.50 m/s²
Time (t) = 70.0 s

d₃ = (22.4)(70.0) + (1/2)(-3.50)(70.0)²
d₃ = 1,568 + (0.5)(-3.50)(4,900)
d₃ = 1,568 + (-8,575)
d₃ = -7,007 m

Total distance covered by the train:
Total distance = d₁ + d₂ + d₃
Total distance = 313.6 + 1,568 + (-7,007)
Total distance = -5,125.4 m

Therefore, the total distance covered by the train is approximately -5,125.4 meters.

a) To sketch the position vs. time graph, we'll break the journey into three segments: acceleration, constant speed, and deceleration.

Segment 1: Acceleration
During this segment, the train starts from rest and accelerates at a rate of 1.60 m/s2 for 14.0s. The position vs. time graph will be a curved line with increasing slope.

Segment 2: Constant Speed
After accelerating, the train runs at a constant speed for 70.0s. The position vs. time graph will be a straight horizontal line.

Segment 3: Deceleration
The train then slows down at a rate of -3.50 m/s2 until it stops at the next station. The position vs. time graph will be a curved line with decreasing slope.

b) To sketch the velocity vs. time graph, we can use the information from the position vs. time graph.

Segment 1: Acceleration
The velocity vs. time graph will be a straight line with a positive slope, representing the acceleration.

Segment 2: Constant Speed
The velocity remains constant, so the graph will be a straight horizontal line.

Segment 3: Deceleration
The velocity vs. time graph will be a straight line with a negative slope, representing the deceleration.

c) To find the total distance covered by the train, we can calculate the distance traveled in each segment and then sum them up.

Segment 1: Acceleration
We can use the equation s = ut + (1/2)at^2, where u is the initial velocity (0), a is the acceleration (1.60 m/s2), and t is the time (14.0s).
s1 = 0(14.0) + (1/2)(1.60)(14.0)^2
s1 = 11.20 m

Segment 2: Constant Speed
The distance traveled during constant speed can be calculated using the equation s = vt, where v is the constant speed and t is the time.
s2 = (velocity)(time)
s2 = (constant speed)(70.0)
s2 = constant speed * 70.0

Segment 3: Deceleration
Using the same equation s = ut + (1/2)at^2, s3 can be calculated by substituting the appropriate values: u = constant speed, a = -3.50 m/s2 (negative due to deceleration), and t = time taken to stop (unknown).
s3 = (constant speed)(t) + (1/2)(-3.50)(t)^2
0 = (constant speed)(t) + (1/2)(-3.50)(t)^2

You would need the value of the constant speed to solve for t and determine s3.

a) To sketch a position vs. time graph, we need to analyze the motion of the subway train in each interval separately and then combine them:

1. The train starts from rest and accelerates at a rate of 1.60 m/s^2 for 14.0 s. In this interval, the position vs. time graph would be a curved line with an increasing slope because the train is accelerating.

2. After accelerating, the train runs at a constant speed for 70.0 s. In this interval, the position vs. time graph would be a straight line with a positive slope because the train covers the same distance in equal time intervals.

3. Finally, the train slows down at a rate of -3.50 m/s^2 until it stops at the next station. In this interval, the position vs. time graph would be a curved line with a decreasing slope because the train is decelerating.

b) To sketch a velocity vs. time graph, we can use the same analysis as above:

1. The train starts from rest and accelerates at a rate of 1.60 m/s^2 for 14.0 s. In this interval, the velocity vs. time graph would be a straight line with a positive slope, indicating that the velocity is increasing.

2. After accelerating, the train runs at a constant speed for 70.0 s. In this interval, the velocity vs. time graph would be a horizontal straight line at the constant velocity.

3. Finally, the train slows down at a rate of -3.50 m/s^2 until it stops at the next station. In this interval, the velocity vs. time graph would be a straight line with a negative slope, indicating that the velocity is decreasing.

c) To find the total distance covered by the train, we need to calculate the distance traveled during each interval and then add them together:

1. Distance during acceleration:
Using the equation of motion, s = ut + (1/2)at^2, where u is the initial velocity (0 m/s), a is the acceleration (1.60 m/s^2), and t is the time (14.0 s), we can calculate the distance traveled during acceleration. Plugging in the values, we get s = (0)(14.0) + (0.5)(1.60)(14.0)^2 = 156.8 m.

2. Distance during constant speed:
The distance traveled during constant speed can be calculated using the formula s = vt, where v is the velocity (which is constant) and t is the time. Plugging in the values, we get s = (v)(70.0). Since the velocity is constant, v = (1.60)(14.0), which equals 22.4 m/s. Therefore, s = (22.4)(70.0) = 1568 m.

3. Distance during deceleration:
Using the same equation as in the first step, but with the acceleration now being -3.50 m/s^2 and the time being 14.0 s, we can calculate the distance traveled during deceleration. Plugging in the values, we get s = (0)(14.0) + (0.5)(-3.50)(14.0)^2 = -343.0 m (the negative sign indicates the direction of deceleration).

Therefore, the total distance covered by the train is the sum of the distances calculated in each interval: 156.8 m + 1568 m - 343.0 m = 1381.8 m.

c. d1 = 0.5a*t^2 = o.8*14^2 = 156.8 m.

V = a*t = 1.6 * 14 = 22.4 m/s.
d2 = 22.4m/s * 70s = 1568 m.

V^2 = Vo^2 + 2ad.
d3 = (V^2-Vo^2)/2a.
d3 = (0-(22.4)^2)/-7 = 71.7 m.

Dt=d1+d2+d3 = 156.8+1568+71.7 = 1797 m.