2020 A/L Physics Past Paper · Question 04 (English Medium)

Answer: (3) $2.5\ \text{m s}^{-1}$

What the question asks

What is asked: the average velocity from $t = 0$ to $t = 4$ s — total displacement ÷ total time.

Reading the options

The options run from $1.5$ to $3.3\ \text{m s}^{-1}$.

The average velocity must lie between the two speeds ($1.0$, $3.0$).

But the object spends more time at $3.0\ \text{m s}^{-1}$, so it should be closer to $3.0$.

Relevant theory

The area between the curve and the time axis = Displacement

displacement = velocity x time

Required equations

$s = v\,t \;(\text{rectangle area})$

$\bar v = \frac{s_1 + s_2}{t}$

Reasoning and calculation

Axes: the horizontal axis is time $t$ (s), the vertical axis is velocity $v$ ($\text{m s}^{-1}$). $So\,the\,area\,under\,the\,graph\,(m\,s^{-1} \times s)\,is\,a\,displacement\,in\,metres$.

Key points: from $0$–$3.0$ s, $v = 3.0\ \text{m s}^{-1}$ (flat line); at $t = 3.0$ s the velocity drops suddenly to $1.0\ \text{m s}^{-1}$; from $3.0$–$4.0$ s, $v = 1.0\ \text{m s}^{-1}$.

$9.0$

$s_1 = 3.0 \times 3.0$

$1.0$

$s_2 = 1.0 \times (4.0 - 3.0)$

$10.0$

$s = 9.0 + 1.0$

$\bar v = \frac{10.0}{4.0}$

$2.5$

$\bar v = 2.5$

Checks

Unit check: area $\text{m s}^{-1} \times \text{s} = \text{m}$; dividing by $\text{s}$ gives $\text{m s}^{-1}$ ✓.

$(v1 * t1 + v2 * (t2 - t1)) / t2$

Second route (time-weighted average): $\bar v = \dfrac{3}{4} \times 3.0 + \dfrac{1}{4} \times 1.0 = 2.25 + 0.25 = 2.5\ \text{m s}^{-1}$ — the same answer.

Conclusion

$\bar v = \frac{9 + 1}{4} = 2.5\ \text{m s}^{-1}$

Total displacement $10$ m, total time $4$ s; average velocity $2.5\ \text{m s}^{-1}$ — the answer is (3).

Taking the plain average of the two speeds $\left(\dfrac{3.0 + 1.0}{2}\right)$.

A plain average does not work when the time spent at each speed differs; divide the total area under the graph by the total time instead.