CAIE M1 2024 November — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with algebraic unknowns
DifficultyModerate -0.8 This is a straightforward mechanics question testing basic kinematics with variable acceleration. Part (a) is simple integration of acceleration, parts (b-d) involve standard velocity-time graph sketching and area calculations. All techniques are routine for M1 level with no novel problem-solving required, making it easier than average.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration
A particle moves in a straight line. It starts from rest, at time \(t = 0\), and accelerates at 0.6 t ms\(^{-2}\) for 4 s, reaching a speed of \(V\) ms\(^{-1}\). The particle then travels at \(V\) ms\(^{-1}\) for 11 s, and finally slows down, with constant deceleration, stopping after a further 5 s.
  1. Show that \(V = 4.8\). [1]
  2. Sketch a velocity-time graph for the motion. [3]
  3. Find an expression, in terms of \(t\), for the velocity of the particle for \(15 \leqslant t \leqslant 20\). [2]
  4. Find the total distance travelled by the particle. [4]
Question 6:
AnswerMarks
6(a)0.6
v=0.6t dt = t2=0.3t2+0 
 
2
AnswerMarks Guidance
t=4  V =0.342 or 0.316=4.8B1 0.6
AG Must see t2 or 0.3t2 and 4 must be actually shown
2
substituted. Merely stating t=4 is not enough to score this mark.
1
Do not allow use of s=ut+ at2 which leads to 4.8 if a=0.6 is
2
used.
1
AnswerMarks Guidance
6(b)Quadratic with correct curvature starting from (0, 0) to (4, 4.8). B1
4
3
2
1
t
2 4 6 8 10 12 14 16 18 20
Note: the grid is for reference – not shown in QP.
Their graph does not need to be to scale.
AnswerMarks Guidance
Horizontal line at from (4, 4.8) to (15, 4.8)B1 The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8.
AnswerMarks Guidance
Line from (15, 4.8) to (20, 0)B1 The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8. Allow all 3
marks if using V instead of 4.8. ISW any extra out of the range for
t of 0 to 20.
If no marks scored then SC B1 for a correct shaped graph with no
numbers.
If using a value of v4.8, allow SC B1 for the first section
correct and SC B1 for the second and third both correct.
3
v
t
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(c)Attempt to find acceleration M1
For calculation oe =−0.96  Allow +0.96.
20−15
AnswerMarks Guidance
v= 19.2−0.96tA1 v= 4.8−0.96(t−15)
Oe e.g. Allow .
2
AnswerMarks
6(d) 4  0.3
 0.3t2dt t3=0.1t3
   
AnswerMarks Guidance
 0  3*M1 Attempt to integrate their v from part (a) provided this came from
integration, but allow a restart here. The power of t must increase
by 1 with a change of coefficient. Use of s=vt scores M0. No
need for limits.
If no integration seen allow SCM1 for answer of 6.4 in place of
M1M1.
0.143−0.103
AnswerMarks Guidance
 DM1 Correct use of correct limit(s) (expect 6.4).
4.811+0.54.85
AnswerMarks Guidance
=52.8+12=64.8 B1 Both correct and added. May be done in one go using a trapezium
(11+16)4.8
.
2
Could do the last stage by integration.
Maximum B1 for final answer 74.4 from thinking the first section
is also straight.
AnswerMarks
Distance = 71.2 mA1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | 0.6
v=0.6t dt = t2=0.3t2+0 
 
2
t=4  V =0.342 or 0.316=4.8 | B1 | 0.6
AG Must see t2 or 0.3t2 and 4 must be actually shown
2
substituted. Merely stating t=4 is not enough to score this mark.
1
Do not allow use of s=ut+ at2 which leads to 4.8 if a=0.6 is
2
used.
1
--- 6(b) ---
6(b) | Quadratic with correct curvature starting from (0, 0) to (4, 4.8). | B1 | 5v
4
3
2
1
t
2 4 6 8 10 12 14 16 18 20
Note: the grid is for reference – not shown in QP.
Their graph does not need to be to scale.
Horizontal line at from (4, 4.8) to (15, 4.8) | B1 | The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8.
Line from (15, 4.8) to (20, 0) | B1 | The points should be specified somehow, but for an accurate
sketch allow a line just below 5 without specifying 4.8. Allow all 3
marks if using V instead of 4.8. ISW any extra out of the range for
t of 0 to 20.
If no marks scored then SC B1 for a correct shaped graph with no
numbers.
If using a value of v4.8, allow SC B1 for the first section
correct and SC B1 for the second and third both correct.
3
v
t
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | Attempt to find acceleration | M1 | 0−4.8
For calculation oe =−0.96  Allow +0.96.
20−15
v= 19.2−0.96t | A1 | v= 4.8−0.96(t−15)
Oe e.g. Allow .
2
--- 6(d) ---
6(d) |  4  0.3
 0.3t2dt t3=0.1t3
   
 0  3 | *M1 | Attempt to integrate their v from part (a) provided this came from
integration, but allow a restart here. The power of t must increase
by 1 with a change of coefficient. Use of s=vt scores M0. No
need for limits.
If no integration seen allow SCM1 for answer of 6.4 in place of
M1M1.
0.143−0.103
  | DM1 | Correct use of correct limit(s) (expect 6.4).
4.811+0.54.85
=52.8+12=64.8  | B1 | Both correct and added. May be done in one go using a trapezium
(11+16)4.8
.
2
Could do the last stage by integration.
Maximum B1 for final answer 74.4 from thinking the first section
is also straight.
Distance = 71.2 m | A1
4
Question | Answer | Marks | Guidance
A particle moves in a straight line. It starts from rest, at time $t = 0$, and accelerates at 0.6 t ms$^{-2}$ for 4 s, reaching a speed of $V$ ms$^{-1}$. The particle then travels at $V$ ms$^{-1}$ for 11 s, and finally slows down, with constant deceleration, stopping after a further 5 s.

\begin{enumerate}[label=(\alph*)]
\item Show that $V = 4.8$. [1]

\item Sketch a velocity-time graph for the motion. [3]

\item Find an expression, in terms of $t$, for the velocity of the particle for $15 \leqslant t \leqslant 20$. [2]

\item Find the total distance travelled by the particle. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q6 [10]}}
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