CAIE M1 2024 June — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeDisplacement from velocity by integration
DifficultyModerate -0.3 This is a straightforward variable acceleration question requiring standard calculus operations: differentiate to find acceleration and set it positive (part a), integrate velocity and solve for displacement = 0 (part b). Both parts use routine A-level mechanics techniques with no conceptual challenges, making it slightly easier than average but not trivial due to the quadratic manipulations required.
Spec3.02f Non-uniform acceleration: using differentiation and integration
A particle \(P\) moves in a straight line. At time \(t\) s after leaving a point \(O\) on the line, \(P\) has velocity \(v\text{ ms}^{-1}\), where \(v = 44t - 6t^2 - 36\).
  1. Find the set of values of \(t\) for which the acceleration of the particle is positive. [2]
  2. Find the two values of \(t\) at which \(P\) returns to \(O\). [3]
Question 2:
AnswerMarks Guidance
2(a)Attempt to differentiate given v M1
one term (which must be the same term); allow
unsimplified.
v
Use of a scores M0.
t
11
4412t 0  t
AnswerMarks Guidance
3A1 44 2
OE, e.g. , 3 , 3.67 or better.
12 3
11
Do not allow t .
3
11
May solve 4412t 0, but final answer must be t .
3
If a lower limit included it must be 0. Allow t 0 or t0
 11  11
. Allow  0,  or  0, .
 3   3 
Alternative Method for Question 2(a):
  11 2 
Use completing the square to get 6 t  
AnswerMarks Guidance
  3  (M1) OE
11
t
AnswerMarks Guidance
3(A1) CWO
If a lower limit included it must be 0. Allow t 0 or t0
 11  11
. Allow  0,  or  0, .
 3   3 
AnswerMarks Guidance
QuestionAnswer Marks
2(a)Alternative Method 2 for Question 2(a):
t t
Solving 44t6t2 360 and find 1 2 , or equivalent.
AnswerMarks Guidance
2(M1) Complete method for finding the value of t at maximum,
b
or use  with correct a and b.
2a
1167
For reference, t  .
3
11
t
AnswerMarks Guidance
3(A1) If a lower limit included it must be 0. Allow t 0 or
 11  11
t0. Allow  0,  or  0, .
 3   3 
2
AnswerMarks Guidance
2(b)Attempt to integrate given v M1
one term (which must be the same term). Use ofsvt is
M0.
44 6
s t11 t2136t c22t2 2t3 36tc
AnswerMarks Guidance
11 21A1 Allow unsimplified.
22t2 2t3 36t 0  t2, 9 and 0 ONLY
AnswerMarks Guidance
 A1 CWO
Ignore t 0 if not rejected.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | Attempt to differentiate given v | M1 | Decrease power by 1 and a change in coefficient in at least
one term (which must be the same term); allow
unsimplified.
v
Use of a scores M0.
t
11
4412t 0  t
3 | A1 | 44 2
OE, e.g. , 3 , 3.67 or better.
12 3
11
Do not allow t .
3
11
May solve 4412t 0, but final answer must be t .
3
If a lower limit included it must be 0. Allow t 0 or t0
 11  11
. Allow  0,  or  0, .
 3   3 
Alternative Method for Question 2(a):
  11 2 
Use completing the square to get 6 t  
  3   | (M1) | OE
11
t
3 | (A1) | CWO
If a lower limit included it must be 0. Allow t 0 or t0
 11  11
. Allow  0,  or  0, .
 3   3 
Question | Answer | Marks | Guidance
2(a) | Alternative Method 2 for Question 2(a):
t t
Solving 44t6t2 360 and find 1 2 , or equivalent.
2 | (M1) | Complete method for finding the value of t at maximum,
b
or use  with correct a and b.
2a
1167
For reference, t  .
3
11
t
3 | (A1) | If a lower limit included it must be 0. Allow t 0 or
 11  11
t0. Allow  0,  or  0, .
 3   3 
2
--- 2(b) ---
2(b) | Attempt to integrate given v | M1 | Increase power by 1 and a change in coefficient in at least
one term (which must be the same term). Use ofsvt is
M0.
44 6
s t11 t2136t c22t2 2t3 36tc
11 21 | A1 | Allow unsimplified.
22t2 2t3 36t 0  t2, 9 and 0 ONLY
  | A1 | CWO
Ignore t 0 if not rejected.
3
Question | Answer | Marks | Guidance
A particle $P$ moves in a straight line. At time $t$ s after leaving a point $O$ on the line, $P$ has velocity $v\text{ ms}^{-1}$, where $v = 44t - 6t^2 - 36$.

\begin{enumerate}[label=(\alph*)]
\item Find the set of values of $t$ for which the acceleration of the particle is positive. [2]

\item Find the two values of $t$ at which $P$ returns to $O$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q2 [5]}}
This paper (7 questions)
View full paper
Q1 3 Q2 5 Q3 6 Q4 7 Q5 8 Q6 11 Q7 10