Easy -2.0 This is a straightforward interpretation question requiring only basic understanding that velocity-time graphs show velocity on the y-axis, zero velocity means stationary, negative gradient means deceleration, area gives displacement, and speed is the magnitude of velocity. No calculations needed, just reading the graph correctly.
The diagram below shows a velocity-time graph for a particle moving with velocity \(v \text{ m s}^{-1}\) at time \(t\) seconds.
\includegraphics{figure_10}
Which statement is correct?
Tick (\(\checkmark\)) one box.
[1 mark]
The particle was stationary for \(9 \leq t \leq 12\)
The particle was decelerating for \(12 \leq t \leq 20\)
The particle had a displacement of zero when \(t = 6\)
The particle's speed when \(t = 4\) was \(-12 \text{ m s}^{-1}\)
The diagram below shows a velocity-time graph for a particle moving with velocity $v \text{ m s}^{-1}$ at time $t$ seconds.
\includegraphics{figure_10}
Which statement is correct?
Tick ($\checkmark$) one box.
[1 mark]
The particle was stationary for $9 \leq t \leq 12$
The particle was decelerating for $12 \leq t \leq 20$
The particle had a displacement of zero when $t = 6$
The particle's speed when $t = 4$ was $-12 \text{ m s}^{-1}$
\hfill \mbox{\textit{AQA Paper 2 2019 Q10 [1]}}