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The displacement-time graph shows a person walking along a road. A) Calculate the average velocity of the whole journey. B) Calculate the instantaneous velocity at point B. C) Calculate the average velocity fro t = 7 to t = 15 seconds.

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Here are the solutions to the displacement-time graph questions:

A) Calculate the average velocity of the whole journey.
Average velocity is defined as the total displacement divided by the total time taken.

  1. Identify the initial and final displacement and time from the graph.

    • Initial displacement at t=0 st=0 \text{ s} is 0 m0 \text{ m}.
    • Final displacement at t=20 st=20 \text{ s} is 0 m0 \text{ m}.
    • Total time is 20 s20 \text{ s}.

  2. Calculate the total displacement and average velocity.

    Total Displacement=Final Displacement−Initial Displacement=0 m−0 m=0 m\text{Total Displacement} = \text{Final Displacement} - \text{Initial Displacement} = 0 \text{ m} - 0 \text{ m} = 0 \text{ m} Average Velocity=Total DisplacementTotal Time=0 m20 s=0 m/s\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{0 \text{ m}}{20 \text{ s}} = 0 \text{ m/s}

    The average velocity of the whole journey is 0 m/s\boxed{\text{0 m/s}}.

    B) Calculate the instantaneous velocity at point B.
    Instantaneous velocity on a displacement-time graph is the gradient (slope) of the graph at that specific point. Point B is located at (16 s,−20 m)(16 \text{ s}, -20 \text{ m}). This point lies on the straight line segment from t=14 st=14 \text{ s} to t=20 st=20 \text{ s}.

  3. Identify two points on the straight line segment that includes point B.

    • Point 1: (t1,d1)=(14 s,−30 m)(t_1, d_1) = (14 \text{ s}, -30 \text{ m})
    • Point 2: (t2,d2)=(20 s,0 m)(t_2, d_2) = (20 \text{ s}, 0 \text{ m})

  4. Calculate the gradient (velocity) using these two points.

    Velocity=ΔdΔt=d2−d1t2−t1=0 m−(−30 m)20 s−14 s=30 m6 s=5 m/s\text{Velocity} = \frac{\Delta d}{\Delta t} = \frac{d_2 - d_1}{t_2 - t_1} = \frac{0 \text{ m} - (-30 \text{ m})}{20 \text{ s} - 14 \text{ s}} = \frac{30 \text{ m}}{6 \text{ s}} = 5 \text{ m/s}

    The instantaneous velocity at point B is 5 m/s\boxed{\text{5 m/s}}.

    C) Calculate the average velocity from t=7 st=7 \text{ s} to t=15 st=15 \text{ s}.
    Average velocity is the total displacement over the given time interval.

  5. Find the displacement at t=7 st=7 \text{ s} and t=15 st=15 \text{ s} from the graph.

    • At t=7 st=7 \text{ s}, the displacement is 20 m20 \text{ m}.
    • At t=15 st=15 \text{ s}, the point is on the segment from (14 s,−30 m)(14 \text{ s}, -30 \text{ m}) to (20 s,0 m)(20 \text{ s}, 0 \text{ m}). The velocity of this segment is 5 m/s5 \text{ m/s} (from part B). Displacement at t=15 s=Displacement at t=14 s+(velocity×Δt)t=15 \text{ s} = \text{Displacement at } t=14 \text{ s} + (\text{velocity} \times \Delta t) d15s=−30 m+(5 m/s×(15 s−14 s))=−30 m+(5 m/s×1 s)=−30 m+5 m=−25 md_{15s} = -30 \text{ m} + (5 \text{ m/s} \times (15 \text{ s} - 14 \text{ s})) = -30 \text{ m} + (5 \text{ m/s} \times 1 \text{ s}) = -30 \text{ m} + 5 \text{ m} = -25 \text{ m}

  6. Calculate the total displacement and average velocity for the interval.

    Total Displacement=d15s−d7s=−25 m−20 m=−45 m\text{Total Displacement} = d_{15s} - d_{7s} = -25 \text{ m} - 20 \text{ m} = -45 \text{ m} Total Time=15 s−7 s=8 s\text{Total Time} = 15 \text{ s} - 7 \text{ s} = 8 \text{ s} Average Velocity=Total DisplacementTotal Time=−45 m8 s=−5.625 m/s\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{-45 \text{ m}}{8 \text{ s}} = -5.625 \text{ m/s}

    The average velocity from t=7 st=7 \text{ s} to t=15 st=15 \text{ s} is -5.625 m/s\boxed{\text{-5.625 m/s}}.

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