CAIE M1 2023 March — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionMarch
Marks5
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring integration of a = 4t² to find v, then integration again to find s. Part (a) is direct substitution after one integration. Part (b) requires setting v = s and solving a simple equation. The polynomial form makes integration routine, and the problem-solving element is minimal—just applying standard kinematics integration twice and solving an algebraic equation.
Spec3.02f Non-uniform acceleration: using differentiation and integration
A particle moves in a straight line starting from rest from a point \(O\). The acceleration of the particle at time \(t\) s after leaving \(O\) is \(a\,\text{m}\,\text{s}^{-2}\), where \(a = 4t^2\).
  1. Find the speed of the particle when \(t = 9\). [2]
  2. Find the time after leaving \(O\) at which the speed (in metres per second) and the distance travelled (in metres) are numerically equal. [3]
Question 3:
AnswerMarks
3(a)Attempt to integrate
 3 3 
4 8
 (v=) t2 = t2 (+c) 
AnswerMarks Guidance
 1.5 3 M1 Increasing power by 1 and a change in coefficient in at least one
term; may be unsimplified.
v=at M0.
AnswerMarks Guidance
Substitute t=9 to get speed = 72 m s–1A1 Or use limits t=0 and t=9.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(b)Attempt at integration of their v
 8 5 
(s=)= 3 t2 = 16 t2 (+c')
 2.5 15 
AnswerMarks Guidance
 *M1 Increasing power by 1 and a change in coefficient in at least one
term; may be unsimplified.
s=vt M0
Their v, which has come from integration in part (a).
Equate their v and their s and attempt to solve for t
 5 3 
16 8 16 8
 t2 = t2  t− =0
AnswerMarks Guidance
 15 3 15 3 DM1 Their v must have come from integration.
Allow if their c from (a) is not 0.
5
time= s
AnswerMarks Guidance
2A1 5
Must discard t=0 and t=− .
2
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | Attempt to integrate
 3 3 
4 8
 (v=) t2 = t2 (+c) 
 1.5 3  | M1 | Increasing power by 1 and a change in coefficient in at least one
term; may be unsimplified.
v=at M0.
Substitute t=9 to get speed = 72 m s–1 | A1 | Or use limits t=0 and t=9.
2
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | Attempt at integration of their v
 8 5 
(s=)= 3 t2 = 16 t2 (+c')
 2.5 15 
  | *M1 | Increasing power by 1 and a change in coefficient in at least one
term; may be unsimplified.
s=vt M0
Their v, which has come from integration in part (a).
Equate their v and their s and attempt to solve for t
 5 3 
16 8 16 8
 t2 = t2  t− =0
 15 3 15 3  | DM1 | Their v must have come from integration.
Allow if their c from (a) is not 0.
5
time= s
2 | A1 | 5
Must discard t=0 and t=− .
2
3
Question | Answer | Marks | Guidance
A particle moves in a straight line starting from rest from a point $O$. The acceleration of the particle at time $t$ s after leaving $O$ is $a\,\text{m}\,\text{s}^{-2}$, where $a = 4t^2$.

\begin{enumerate}[label=(\alph*)]
\item Find the speed of the particle when $t = 9$. [2]

\item Find the time after leaving $O$ at which the speed (in metres per second) and the distance travelled (in metres) are numerically equal. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q3 [5]}}
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