Easy -1.2 This is a straightforward one-mark question requiring direct application of the formula P = Fv at maximum speed (where driving force equals resistance). It involves simple rearrangement and calculation with no problem-solving or conceptual challenge, making it easier than average.
2 A train is travelling at maximum speed with its engine using its maximum power of 1800 kW
When travelling at this speed the train experiences a total resistive force of 40000 N Find the maximum speed of the train.
Circle your answer. [0pt]
[1 mark]
\(22 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
\(45 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
\(54 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
\(90 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
2 A train is travelling at maximum speed with its engine using its maximum power of 1800 kW
When travelling at this speed the train experiences a total resistive force of 40000 N Find the maximum speed of the train.
Circle your answer.\\[0pt]
[1 mark]\\
$22 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
$45 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
$54 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
$90 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2018 Q2 [1]}}