Easy -1.2 This question tests basic understanding that displacement equals the area under a velocity-time graph. Students need only identify which graph has equal positive and negative areas over 4 seconds—a straightforward conceptual check requiring minimal calculation.
12 A particle moves in a straight line.
After the first 4 seconds of its motion, the displacement of the particle from its initial position is 0 metres.
One of the graphs on the opposite page shows the velocity \(v \mathrm {~ms} ^ { - 1 }\) of the particle after time \(t\) seconds of its motion.
Identify the correct graph.
Tick ( \(\checkmark\) ) one box.
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-19_2249_896_260_484}
Graph showing \(v\)-\(t\) relationship: starts at \(v=5\), decreases linearly to \(v=-5\) at \(t=2\), then increases linearly back to \(v=5\) at \(t=4\)
B1
Ticks correct box
## Question 12:
Graph showing $v$-$t$ relationship: starts at $v=5$, decreases linearly to $v=-5$ at $t=2$, then increases linearly back to $v=5$ at $t=4$ | B1 | Ticks correct box
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12 A particle moves in a straight line.\\
After the first 4 seconds of its motion, the displacement of the particle from its initial position is 0 metres.
One of the graphs on the opposite page shows the velocity $v \mathrm {~ms} ^ { - 1 }$ of the particle after time $t$ seconds of its motion.
Identify the correct graph.\\
Tick ( $\checkmark$ ) one box.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-19_2249_896_260_484}
\hfill \mbox{\textit{AQA Paper 2 2023 Q12 [1]}}