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The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while the cyclist is moving with constant velocity, and (ii) when the velocity of the cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
Show Answer
Shading: the rectangular region under the flat part is the displacement during constant velocity; the region under the down-sloping part is the displacement while velocity is decreasing.
Displacement = area under the whole graph. Reading Fig. 4.30 (ramp up 0–20 s to 3 m s⁻¹, constant 3 m s⁻¹ for 20–100 s, then falling to about 2 m s⁻¹ by 120 s):
\( s \approx \tfrac{1}{2}(20)(3) + (80)(3) + \tfrac{1}{2}(3+2)(20) = 30 + 240 + 50 = 320\ \text{m} \).
Average acceleration \(= \dfrac{v_{120} - v_{0}}{120} = \dfrac{2 - 0}{120} \approx 0.017\ \text{m s}^{-2}\). (Read the exact values from your copy of Fig. 4.30.)
