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The following table provides the displacement of a car during a period of 10 seconds. Initially, the car moves in a northerly direction.

Physics

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7 parts
1.2

1.875 m/s2\text{1.875 m/s}^2

1.3

787.5 m\text{787.5 m}

2.1

Car L\text{Car L}

2.2

16 m\text{16 m}

3.2

7 m/s\text{7 m/s}

3.3

0 m/s\text{0 m/s}

3.4

-4 m/s\text{-4 m/s}

Step-by-step solution

14 steps

Part 1.1

Describe the motion of the train.

The train's motion can be described in five segments:

  • From t=0t = 0 s to t=15t = 15 s, the train accelerates uniformly from rest to a velocity of 15 m/s15 \text{ m/s}.
  • From t=15t = 15 s to t=20t = 20 s, the train moves at a constant velocity of 15 m/s15 \text{ m/s}.
  • From t=20t = 20 s to t=28t = 28 s, the train accelerates uniformly from 15 m/s15 \text{ m/s} to 30 m/s30 \text{ m/s}.
  • From t=28t = 28 s to t=40t = 40 s, the train moves at a constant velocity of 30 m/s30 \text{ m/s}.
  • From t=40t = 40 s to t=44t = 44 s, the train decelerates uniformly from 30 m/s30 \text{ m/s} to rest.

Part 1.2

1.875 m/s2\text{1.875 m/s}^2

Calculate the acceleration of the train between 20 and 28 s.

  1. Identify initial and final velocities and times for the interval.

    At ti=20 st_i = 20 \text{ s}, vi=15 m/sv_i = 15 \text{ m/s}.
    At tf=28 st_f = 28 \text{ s}, vf=30 m/sv_f = 30 \text{ m/s}.

  2. Use the acceleration formula.

    a=vf−vitf−tia = \frac{v_f - v_i}{t_f - t_i} a=30 m/s−15 m/s28 s−20 sa = \frac{30 \text{ m/s} - 15 \text{ m/s}}{28 \text{ s} - 20 \text{ s}} a=15 m/s8 sa = \frac{15 \text{ m/s}}{8 \text{ s}} a=1.875 m/s2a = 1.875 \text{ m/s}^2

    The acceleration of the train between 20 and 28 s is 1.875 m/s2\boxed{\text{1.875 m/s}^2}.

Part 1.3

787.5 m\text{787.5 m}

Calculate the distance travelled in the 44 s.

The distance travelled is the total area under the velocity-time graph.

  1. Calculate the area for each segment.

    • Segment 1 (0-15 s, triangle):
    A1=12×base×height=12×15 s×15 m/s=112.5 mA_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \text{ s} \times 15 \text{ m/s} = 112.5 \text{ m}
    • Segment 2 (15-20 s, rectangle):
    A2=width×height=(20 s−15 s)×15 m/s=5 s×15 m/s=75 mA_2 = \text{width} \times \text{height} = (20 \text{ s} - 15 \text{ s}) \times 15 \text{ m/s} = 5 \text{ s} \times 15 \text{ m/s} = 75 \text{ m}
    • Segment 3 (20-28 s, trapezium):
    A3=12×(sum of parallel sides)×height=12×(15 m/s+30 m/s)×(28 s−20 s)A_3 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (15 \text{ m/s} + 30 \text{ m/s}) \times (28 \text{ s} - 20 \text{ s}) A3=12×45 m/s×8 s=180 mA_3 = \frac{1}{2} \times 45 \text{ m/s} \times 8 \text{ s} = 180 \text{ m}
    • Segment 4 (28-40 s, rectangle):
    A4=width×height=(40 s−28 s)×30 m/s=12 s×30 m/s=360 mA_4 = \text{width} \times \text{height} = (40 \text{ s} - 28 \text{ s}) \times 30 \text{ m/s} = 12 \text{ s} \times 30 \text{ m/s} = 360 \text{ m}
    • Segment 5 (40-44 s, triangle):
    A5=12×base×height=12×(44 s−40 s)×30 m/s=12×4 s×30 m/s=60 mA_5 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (44 \text{ s} - 40 \text{ s}) \times 30 \text{ m/s} = \frac{1}{2} \times 4 \text{ s} \times 30 \text{ m/s} = 60 \text{ m}

  2. Sum all the areas to find the total distance.

    Total Distance=A1+A2+A3+A4+A5\text{Total Distance} = A_1 + A_2 + A_3 + A_4 + A_5 Total Distance=112.5 m+75 m+180 m+360 m+60 m\text{Total Distance} = 112.5 \text{ m} + 75 \text{ m} + 180 \text{ m} + 360 \text{ m} + 60 \text{ m} Total Distance=787.5 m\text{Total Distance} = 787.5 \text{ m}

    The total distance travelled in 44 s is 787.5 m\boxed{\text{787.5 m}}.

Part 1.4

Draw an acceleration-time graph from this velocity-time graph.

To draw the acceleration-time graph, we calculate the acceleration for each segment:

  • From t=0t = 0 s to t=15t = 15 s: a=15 m/s−0 m/s15 s−0 s=1 m/s2a = \frac{15 \text{ m/s} - 0 \text{ m/s}}{15 \text{ s} - 0 \text{ s}} = 1 \text{ m/s}^2.
  • From t=15t = 15 s to t=20t = 20 s: a=15 m/s−15 m/s20 s−15 s=0 m/s2a = \frac{15 \text{ m/s} - 15 \text{ m/s}}{20 \text{ s} - 15 \text{ s}} = 0 \text{ m/s}^2.
  • From t=20t = 20 s to t=28t = 28 s: a=30 m/s−15 m/s28 s−20 s=15 m/s8 s=1.875 m/s2a = \frac{30 \text{ m/s} - 15 \text{ m/s}}{28 \text{ s} - 20 \text{ s}} = \frac{15 \text{ m/s}}{8 \text{ s}} = 1.875 \text{ m/s}^2.
  • From t=28t = 28 s to t=40t = 40 s: a=30 m/s−30 m/s40 s−28 s=0 m/s2a = \frac{30 \text{ m/s} - 30 \text{ m/s}}{40 \text{ s} - 28 \text{ s}} = 0 \text{ m/s}^2.
  • From t=40t = 40 s to t=44t = 44 s: a=0 m/s−30 m/s44 s−40 s=−30 m/s4 s=−7.5 m/s2a = \frac{0 \text{ m/s} - 30 \text{ m/s}}{44 \text{ s} - 40 \text{ s}} = \frac{-30 \text{ m/s}}{4 \text{ s}} = -7.5 \text{ m/s}^2.

The acceleration-time graph would consist of horizontal line segments:

  • A horizontal line at a=1 m/s2a = 1 \text{ m/s}^2 from t=0t = 0 s to t=15t = 15 s.
  • A horizontal line at a=0 m/s2a = 0 \text{ m/s}^2 from t=15t = 15 s to t=20t = 20 s.
  • A horizontal line at a=1.875 m/s2a = 1.875 \text{ m/s}^2 from t=20t = 20 s to t=28t = 28 s.
  • A horizontal line at a=0 m/s2a = 0 \text{ m/s}^2 from t=28t = 28 s to t=40t = 40 s.
  • A horizontal line at a=−7.5 m/s2a = -7.5 \text{ m/s}^2 from t=40t = 40 s to t=44t = 44 s.

Part 2.1

Car L\text{Car L}

Identify the car with the greatest velocity.

  1. Understand that velocity on a position-time graph is represented by the gradient (slope) of the line.

  2. Visually compare the slopes of lines K and L.

    Line L is steeper than line K.
    Car L has a greater gradient, indicating a greater velocity.
    The car with the greatest velocity is Car L\boxed{\text{Car L}}.

Part 2.2

16 m\text{16 m}

How far apart are the two cars at t = 0 s?

  1. Read the initial position of each car at t=0t = 0 s from the graph.

    • Car K starts at x=16 mx = 16 \text{ m}.
    • Car L starts at x=0 mx = 0 \text{ m}.

  2. Calculate the difference in their positions.

    Distance apart=∣xK−xL∣=∣16 m−0 m∣=16 m\text{Distance apart} = |x_K - x_L| = |16 \text{ m} - 0 \text{ m}| = 16 \text{ m}

    The two cars are 16 m\boxed{\text{16 m}} apart at t=0t = 0 s.

Part 3.1

Draw an accurate displacement-time graph to represent the motion of the car.

To draw the graph, plot the given points (Time (s), Displacement (m)) and connect them with straight lines:

  • (0, 0)
  • (2, 14)
  • (4, 28)
  • (6, 28)
  • (8, 20)
  • (10, 12)

The graph would show:

  • A line segment from (0,0) to (2,14).
  • A line segment from (2,14) to (4,28).
  • A horizontal line segment from (4,28) to (6,28).
  • A line segment from (6,28) to (8,20).
  • A line segment from (8,20) to (10,12).

Part 3.2

7 m/s\text{7 m/s}

Use the graph to determine the velocity of the car after 3 seconds.

  1. Identify the time interval that includes t=3t = 3 s.

    This is the interval from t=2t = 2 s to t=4t = 4 s.

  2. Calculate the gradient (velocity) for this segment using the points (2 s, 14 m) and (4 s, 28 m).

    v=ΔxΔt=28 m−14 m4 s−2 sv = \frac{\Delta x}{\Delta t} = \frac{28 \text{ m} - 14 \text{ m}}{4 \text{ s} - 2 \text{ s}} v=14 m2 sv = \frac{14 \text{ m}}{2 \text{ s}} v=7 m/sv = 7 \text{ m/s}

    The velocity of the car after 3 seconds is 7 m/s\boxed{\text{7 m/s}}.

Part 3.3

0 m/s\text{0 m/s}

What was the car's velocity after 5 seconds? Justify your answer.

  1. Identify the time interval that includes t=5t = 5 s.

    This is the interval from t=4t = 4 s to t=6t = 6 s.

  2. Calculate the gradient (velocity) for this segment using the points (4 s, 28 m) and (6 s, 28 m).

    v=ΔxΔt=28 m−28 m6 s−4 sv = \frac{\Delta x}{\Delta t} = \frac{28 \text{ m} - 28 \text{ m}}{6 \text{ s} - 4 \text{ s}} v=0 m2 sv = \frac{0 \text{ m}}{2 \text{ s}} v=0 m/sv = 0 \text{ m/s}

    The car's velocity after 5 seconds was 0 m/s\boxed{\text{0 m/s}}. This is because the displacement remained constant at 28 m28 \text{ m} between t=4 st = 4 \text{ s} and t=6 st = 6 \text{ s}, indicating the car was stationary.

Part 3.4

-4 m/s\text{-4 m/s}

Use the graph to determine the car's velocity after 9 seconds.

  1. Identify the time interval that includes t=9t = 9 s.

    This is the interval from t=8t = 8 s to t=10t = 10 s.

  2. Calculate the gradient (velocity) for this segment using the points (8 s, 20 m) and (10 s, 12 m).

    v=ΔxΔt=12 m−20 m10 s−8 sv = \frac{\Delta x}{\Delta t} = \frac{12 \text{ m} - 20 \text{ m}}{10 \text{ s} - 8 \text{ s}} v=−8 m2 sv = \frac{-8 \text{ m}}{2 \text{ s}} v=−4 m/sv = -4 \text{ m/s}

    The car's velocity after 9 seconds is -4 m/s\boxed{\text{-4 m/s}}.

Part 3.5

In words, describe the motion of the car during the 10 seconds.

The car starts from rest at the origin and moves in a northerly direction, increasing its speed initially. From t=0t = 0 s to t=4t = 4 s, its displacement increases from 0 m0 \text{ m} to 28 m28 \text{ m}. Between t=4t = 4 s and t=6t = 6 s, the car stops and remains stationary at 28 m28 \text{ m} displacement. From t=6t = 6 s to t=10t = 10 s, the car moves back towards its starting point (in the southerly direction), with its displacement decreasing from 28 m28 \text{ m} to 12 m12 \text{ m}.

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