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Kinematics: Question 6

Syllabus 2.1

Multiple choice AS 1 mark

A jogger runs along a straight, level path. Her displacement–time graph for a 70 s70\text{ s} run can be described as follows: from t=0t=0 to t=20 st=20\text{ s}, her displacement increases uniformly from 00 to 100 m100\text{ m}; from t=20 st=20\text{ s} to t=50 st=50\text{ s}, her displacement stays constant at 100 m100\text{ m} while she stops to tie her shoelace; from t=50 st=50\text{ s} to t=70 st=70\text{ s}, her displacement decreases uniformly from 100 m100\text{ m} to 40 m40\text{ m} as she jogs back part of the way.

What is the jogger's velocity during the interval t=50 st=50\text{ s} to t=70 st=70\text{ s}?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Recall what the gradient of a displacement–time graph represents

On a displacement–time graph, the gradient of the line at any instant gives the velocity: a positive gradient means the object is moving in the positive direction, and a negative gradient means it is moving in the negative direction (back towards the start).

Step 2: Identify the change in displacement and time for this segment

For the interval t=50 st=50\text{ s} to t=70 st=70\text{ s}, the displacement changes from 100 m100\text{ m} to 40 m40\text{ m}: Δs=40−100=−60 m\Delta s = 40 - 100 = -60\text{ m} Δt=70−50=20 s\Delta t = 70 - 50 = 20\text{ s}

Step 3: Calculate the gradient (velocity)

v=ΔsΔt=−6020v = \frac{\Delta s}{\Delta t} = \frac{-60}{20} v=−3.0 m s−1v = -3.0\text{ m s}^{-1}

The negative sign shows that the jogger’s velocity is directed opposite to her original direction of travel. That is, back towards her starting point. The magnitude of her velocity during this segment is 3.0 m s−13.0\text{ m s}^{-1}.

Why the other options are wrong

  • B: has the correct magnitude but the wrong direction. The decreasing displacement means she is moving back towards the start, not away from it.
  • C: comes from dividing 60 m60\text{ m} by 10 s10\text{ s} instead of the correct 20 s20\text{ s} duration of this segment, doubling the magnitude.
  • D: comes from dividing 60 m60\text{ m} by the full 70 s70\text{ s} of the run instead of isolating the 20 s20\text{ s} duration of this particular segment.

Final answer

  • Velocity during t=50 st=50\text{ s} to t=70 st=70\text{ s} =3.0 m s−1= \boxed{3.0}\text{ m s}^{-1}, directed back towards her starting point, option A.