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Motion and Motion Graphs: Question 4

Syllabus 1.2

Structured Extended 9 marks

A delivery drone flies in a straight line. Its speed at different times is shown below.

Time, tt / s Speed, vv / m/s Description
00 to 55 0→150 \rightarrow 15 speed increases at a constant rate (straight line on graph)
55 to 1515 1515 (constant) speed stays constant
1515 to 2020 15→015 \rightarrow 0 speed decreases at a constant rate (straight line on graph)

After t=20 st = 20\text{ s}, the drone speeds up again. From t=20 st = 20\text{ s} to t=30 st = 30\text{ s}, the speed–time graph for the drone is a curve that becomes less steep as time goes on (rather than a straight line).

(a) Calculate the acceleration of the drone during the first phase, from t=0t = 0 to t=5 st = 5\text{ s}. [2]

(b) State the acceleration of the drone during the second phase, from t=5 st = 5\text{ s} to t=15 st = 15\text{ s}, and explain your answer. [2]

(c) Calculate the acceleration of the drone during the third phase, from t=15 st = 15\text{ s} to t=20 st = 20\text{ s}. State whether your answer represents an acceleration or a deceleration, and explain how the sign of your answer shows this. [3]

(d) State what the shape of the graph during the fourth phase (from t=20 st = 20\text{ s} to t=30 st = 30\text{ s}) shows about the drone's acceleration, and explain how you can tell this from the graph. [2]

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Worked solution

Part (a): Acceleration during the first phase

Acceleration is the change in velocity per unit time:

a=ΔvΔta = \frac{\Delta v}{\Delta t}

During this phase, the speed changes from 0 m/s0\text{ m/s} to 15 m/s15\text{ m/s} over 5 s5\text{ s}:

a=15 m/s−0 m/s5 s=3.0 m/s2a = \frac{15\text{ m/s} - 0\text{ m/s}}{5\text{ s}} = 3.0\text{ m/s}^2

Part (b): Acceleration during the second phase

During this phase, the drone’s speed stays constant at 15 m/s15\text{ m/s}, so there is no change in velocity:

Δv=15 m/s−15 m/s=0 m/s\Delta v = 15\text{ m/s} - 15\text{ m/s} = 0\text{ m/s}

Using a=ΔvΔta = \dfrac{\Delta v}{\Delta t}, since Δv=0\Delta v = 0, the acceleration is:

a=0 m/s2a = 0\text{ m/s}^2

This makes sense: a straight, horizontal section of a speed-time graph has a gradient of zero, and the gradient of a speed-time graph represents acceleration.

Part (c): Acceleration during the third phase

During this phase, the speed decreases from 15 m/s15\text{ m/s} to 0 m/s0\text{ m/s} over 5 s5\text{ s}:

a=ΔvΔt=0 m/s−15 m/s5 sa = \frac{\Delta v}{\Delta t} = \frac{0\text{ m/s} - 15\text{ m/s}}{5\text{ s}}

a=−3.0 m/s2a = -3.0\text{ m/s}^2

The negative sign shows this is a deceleration: the drone’s velocity is decreasing (it is slowing down), so the acceleration acts in the opposite direction to its motion.

Part (d): The curved fourth phase

Between t=20 st = 20\text{ s} and t=30 st = 30\text{ s}, the graph is a curve, not a straight line, so the gradient of the graph is not constant. Because the curve becomes less steep over time, the gradient, and therefore the acceleration, is decreasing. This shows the drone’s acceleration is changing (non-uniform) during this phase, unlike the constant accelerations of the straight-line sections earlier in the flight.

Final answers

  • (a) Acceleration in the first phase == 3.0 m/s23.0\text{ m/s}^2
  • (b) Acceleration in the second phase == 0 m/s20\text{ m/s}^2 (speed is constant)
  • (c) Acceleration in the third phase == −3.0 m/s2-3.0\text{ m/s}^2, a deceleration
  • (d) The acceleration in the fourth phase is decreasing (changing), shown by the curve flattening