CAIE M1 2022 June — Question 7 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypePiecewise motion functions
DifficultyStandard +0.3 This is a straightforward variable acceleration question requiring differentiation to find accelerations at t=10 from both expressions (showing discontinuity), then integration of two velocity functions to find distances. All techniques are standard M1 procedures with no novel problem-solving required, making it slightly easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration
A particle \(P\) moves in a straight line. The velocity \(v\text{ms}^{-1}\) at time \(t\) seconds is given by $$v = 0.5t \quad \text{for } 0 \leqslant t \leqslant 10,$$ $$v = 0.25t^2 - 8t + 60 \quad \text{for } 10 < t \leqslant 20.$$
  1. Show that there is an instantaneous change in the acceleration of the particle at \(t = 10\). [3]
  2. Find the total distance covered by \(P\) in the interval \(0 \leqslant t \leqslant 20\). [6]
Question 7:
AnswerMarks
7(a)dv
a 0.5
AnswerMarks Guidance
dtB1 For acceleration during the first 10 seconds
dv
Differentiate to get a  20.25t8 0.5t8
AnswerMarks Guidance
dtB1 Allow unsimplified
a 0.51083B1 CWO. Do not award final B mark if more than 2
accelerations seen and not discarded, 2/3 maximum
Ignore any comments, correct or incorrect
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
7(b)Get distance in first 10 seconds as 25 B1
From suvat or from 0.5t dt
0
AnswerMarks Guidance
v0 when t 12 and t 20B1 SOI
Attempt to integrate v
s  0.25t2 8t60  dt
AnswerMarks Guidance
 *M1 For integration, the power of t must increase by 1 in at least
1 term with a change of coefficient in the same term.
s vt is M0
0.25 8  1 
s  t3  t2 60tc  t3 4t2 60t c
 
AnswerMarks Guidance
3 2  12 A1 Allow unsimplified
 1 
Attempt to evaluate their t34t2 60t for t 10 to t 12
 
12 
AnswerMarks Guidance
and t 12 to t 20DM1 Using the correct limits correctly
 850 800  14  64
s   25288   288  25      51 m
AnswerMarks Guidance
 3  3  3  3 A1
QuestionAnswer Marks
7(b)Special Case for those who use a calculator to integrate. Maximum 4/6
Get distance in first 10 seconds as 25B1 10
From suvat or 0.5t dt
0
AnswerMarks Guidance
v0 when t 12 and t 20B1 SOI
12 14
Either s    0.25t2 8t60  dt  4.67
3
10
20 64
Or s    0.25t2 8t60  dt  21.3
3
AnswerMarks Guidance
12B1 20
Allow  0.25t2 8t60 dt 26
10
 14 64
s  25   51 m
 
AnswerMarks Guidance
 3 3 B1 Allow if t 12 and t 20 not found for 3 marks
6
Question 7:
--- 7(a) ---
7(a) | dv
a 0.5
dt | B1 | For acceleration during the first 10 seconds
dv
Differentiate to get a  20.25t8 0.5t8
dt | B1 | Allow unsimplified
a 0.51083 | B1 | CWO. Do not award final B mark if more than 2
accelerations seen and not discarded, 2/3 maximum
Ignore any comments, correct or incorrect
3
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Get distance in first 10 seconds as 25 | B1 | 10
From suvat or from 0.5t dt
0
v0 when t 12 and t 20 | B1 | SOI
Attempt to integrate v
s  0.25t2 8t60  dt
  | *M1 | For integration, the power of t must increase by 1 in at least
1 term with a change of coefficient in the same term.
s vt is M0
0.25 8  1 
s  t3  t2 60tc  t3 4t2 60t c
 
3 2  12  | A1 | Allow unsimplified
 1 
Attempt to evaluate their t34t2 60t for t 10 to t 12
 
12 
and t 12 to t 20 | DM1 | Using the correct limits correctly
 850 800  14  64
s   25288   288  25      51 m
 3  3  3  3  | A1
Question | Answer | Marks | Guidance
7(b) | Special Case for those who use a calculator to integrate. Maximum 4/6
Get distance in first 10 seconds as 25 | B1 | 10
From suvat or 0.5t dt
0
v0 when t 12 and t 20 | B1 | SOI
12 14
Either s    0.25t2 8t60  dt  4.67
3
10
20 64
Or s    0.25t2 8t60  dt  21.3
3
12 | B1 | 20
Allow  0.25t2 8t60 dt 26
10
 14 64
s  25   51 m
 
 3 3  | B1 | Allow if t 12 and t 20 not found for 3 marks
6
A particle $P$ moves in a straight line. The velocity $v\text{ms}^{-1}$ at time $t$ seconds is given by
$$v = 0.5t \quad \text{for } 0 \leqslant t \leqslant 10,$$
$$v = 0.25t^2 - 8t + 60 \quad \text{for } 10 < t \leqslant 20.$$

\begin{enumerate}[label=(\alph*)]
\item Show that there is an instantaneous change in the acceleration of the particle at $t = 10$. [3]

\item Find the total distance covered by $P$ in the interval $0 \leqslant t \leqslant 20$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q7 [9]}}
This paper (7 questions)
View full paper
Q1 5 Q2 6 Q3 7 Q4 9 Q5 6 Q6 8 Q7 9