CAIE M1 Specimen — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
SessionSpecimen
Marks10
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVerifying given motion properties
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: integration to find displacement, differentiation to find acceleration, and solving a quadratic equation. Part (i) involves verification by integration and simple differentiation. Part (ii) requires finding maximum velocity (setting dv/dt=0) then solving a quadratic, which are routine A-level mechanics procedures with no novel problem-solving required.
Spec1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration

A particle \(P\) moves in a straight line, starting from a point \(O\). The velocity of \(P\), measured in m s\(^{-1}\), at time \(t\) s after leaving \(O\) is given by $$v = 0.6t - 0.03t^2.$$
  1. Verify that, when \(t = 5\), the particle is 6.25 m from \(O\). Find the acceleration of the particle at this time. [4]
  2. Find the values of \(t\) at which the particle is travelling at half of its maximum velocity. [6]

Question 6:

AnswerMarks Guidance
6(i)s = 0.3t2 – 0.01t3 1
s(5) = 0.3 × 52 – 0.01 × 53 = 6.251 A1
a = 0.6 – 0.06t1 M1
a(5) = 0.6 – 0.0 × 5 = 0.3 ms–21 A1
4

AnswerMarks Guidance
6(ii)Maximum velocity is when
0.6 – 0.06t = 01 M1
[t = 10]1 M1
Max velocity = 3 ms–11 A1
0.6t – 0.03t2 = 1.5
AnswerMarks Guidance
[t2 – 20t + 50 = 0]1 M1
solve a three term quadratic
AnswerMarks Guidance
Times are 2.93 s1 A1
and 17.07 s1 A1
6
Question 6:
--- 6(i) ---
6(i) | s = 0.3t2 – 0.01t3 | 1 | M1 | For integration
s(5) = 0.3 × 52 – 0.01 × 53 = 6.25 | 1 | A1
a = 0.6 – 0.06t | 1 | M1 | For differentiation
a(5) = 0.6 – 0.0 × 5 = 0.3 ms–2 | 1 | A1
4
--- 6(ii) ---
6(ii) | Maximum velocity is when
0.6 – 0.06t = 0 | 1 | M1 | For setting a = 0
[t = 10] | 1 | M1 | For solving a = 0
Max velocity = 3 ms–1 | 1 | A1
0.6t – 0.03t2 = 1.5
[t2 – 20t + 50 = 0] | 1 | M1 | Setting velocity = half its maximum and attempting to
solve a three term quadratic
Times are 2.93 s | 1 | A1
and 17.07 s | 1 | A1
6
A particle $P$ moves in a straight line, starting from a point $O$. The velocity of $P$, measured in m s$^{-1}$, at time $t$ s after leaving $O$ is given by
$$v = 0.6t - 0.03t^2.$$

\begin{enumerate}[label=(\roman*)]
\item Verify that, when $t = 5$, the particle is 6.25 m from $O$. Find the acceleration of the particle at this time. [4]

\item Find the values of $t$ at which the particle is travelling at half of its maximum velocity. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1  Q6 [10]}}