Standard +0.3 This is a straightforward differential equation problem requiring standard techniques: setting up a = k/v from inverse proportionality, finding k using initial conditions, separating variables (v dv = k dt), and integrating to get v². Part (b) requires solving |v| = |a| which is algebraic substitution. All steps are routine for A-level Further Maths with no novel insight required, making it slightly easier than average.
A train is moving along a straight horizontal track. At time t seconds, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), its acceleration is \(a \mathrm {~ms} ^ { - 2 }\), and \(a\) is inversely proportional to V . At time \(\mathrm { t } = 1\), \(v = 5\) and \(a = 1 \cdot 8\).
a) i) Write down a differential equation satisfied by V.
ii) Show that \(v ^ { 2 } = 18 t + 7\).
b) Find the time at which the magnitude of the velocity is equal to the magnitude of the acceleration.
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A train is moving along a straight horizontal track. At time t seconds, its velocity is $v \mathrm {~ms} ^ { - 1 }$, its acceleration is $a \mathrm {~ms} ^ { - 2 }$, and $a$ is inversely proportional to V . At time $\mathrm { t } = 1$, $v = 5$ and $a = 1 \cdot 8$.
a) i) Write down a differential equation satisfied by V.\\
ii) Show that $v ^ { 2 } = 18 t + 7$.\\
b) Find the time at which the magnitude of the velocity is equal to the magnitude of the acceleration.
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\hfill \mbox{\textit{WJEC Unit 4 2023 Q10}}