AQA Paper 2 2021 June — Question 11 1 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2021
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeApplied rate of change
DifficultyEasy -1.8 This is a straightforward single-step differentiation of an exponential function using the chain rule, presented as a multiple-choice question. It requires only direct application of a standard formula (d/dt of e^(kt) = ke^(kt)) with no problem-solving, making it significantly easier than average A-level questions.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)

11 A particle's displacement, \(r\) metres, with respect to time, \(t\) seconds, is defined by the equation $$r = 3 \mathrm { e } ^ { 0.5 t }$$ Find an expression for the velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of the particle at time \(t\) seconds.
Circle your answer. \(v = 1.5 \mathrm { e } ^ { 0.5 t }\) \(v = 6 \mathrm { e } ^ { 0.5 t }\) \(v = 1.5 t \mathrm { e } ^ { 0.5 t }\) \(v = 6 t e ^ { 0.5 t }\)

Question 11:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(v = 1.5e^{0.5t}\)B1 (AO1.1b) Circles correct answer
## Question 11:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 1.5e^{0.5t}$ | B1 (AO1.1b) | Circles correct answer |

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11 A particle's displacement, $r$ metres, with respect to time, $t$ seconds, is defined by the equation

$$r = 3 \mathrm { e } ^ { 0.5 t }$$

Find an expression for the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the particle at time $t$ seconds.\\
Circle your answer.\\
$v = 1.5 \mathrm { e } ^ { 0.5 t }$\\
$v = 6 \mathrm { e } ^ { 0.5 t }$\\
$v = 1.5 t \mathrm { e } ^ { 0.5 t }$\\
$v = 6 t e ^ { 0.5 t }$

\hfill \mbox{\textit{AQA Paper 2 2021 Q11 [1]}}