CAIE M1 2017 March — Question 5 12 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2017
SessionMarch
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypePiecewise motion functions
DifficultyStandard +0.3 This is a straightforward piecewise velocity function question requiring continuity conditions (part i), basic graph sketching (part ii), and integration to find distance (part iii). All techniques are standard M1 material with clear signposting and no novel problem-solving required—slightly easier than average due to the structured guidance.
Spec1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time
A particle \(P\) moves in a straight line starting from a point \(O\) and comes to rest \(35\) s later. At time \(t\) s after leaving \(O\), the velocity \(v\) m s\(^{-1}\) of \(P\) is given by $$v = \frac{4}{5}t^2 \quad 0 \leq t \leq 5,$$ $$v = 2t + 10 \quad 5 \leq t \leq 15,$$ $$v = a + bt^2 \quad 15 \leq t \leq 35,$$ where \(a\) and \(b\) are constants such that \(a > 0\) and \(b < 0\).
  1. Show that the values of \(a\) and \(b\) are \(49\) and \(-0.04\) respectively. [3]
  2. Sketch the velocity-time graph. [4]
  3. Find the total distance travelled by \(P\) during the \(35\) s. [5]
Question 5:
AnswerMarks Guidance
5(i)0= a + b × 352
40 = a + b × 152M1 For matching velocities at
t = 15 and using v = 0 at t = 35
[1000b = -40 → b = –0.04]
AnswerMarks Guidance
[a = 0.04 × 352 = 49]M1 Solve for a and b
a = 49 and b = -0.04 AGA1
Total:3
AnswerMarks Guidance
5(ii)0 ⩽ t ⩽ 5 correct B1
(5,20), concave up
AnswerMarks Guidance
5 ⩽ t ⩽ 15 correctB1 Line from (5,20) to (15,40)
15 ⩽ t ⩽ 35 correctB1 Decreasing quadratic, from (15,40) to
(35,0), concave down
AnswerMarks
20 and 40 seen correct on v-axisB1
Total:4
AnswerMarks
5(iii)5 100
A = ∫0.8t2dt =
1 3
AnswerMarks
0B1
A = 1( 20+40 )×10=300
AnswerMarks Guidance
2 2M1 Using trapezium rule or integration for
t = 5 to t = 15
35( )
A = ∫ a+bt2 dt
3
15
0.04
=49t− t3
AnswerMarks Guidance
3M1 Attempt to integrate the quadratic
function
from t = 15 to t = 35
A = 453.3333 = 1360/3
AnswerMarks Guidance
3A1
Total Distance = 2360/3 = 787mA1
Total:5
QuestionAnswer Marks 
Question 5:
--- 5(i) ---
5(i) | 0= a + b × 352
40 = a + b × 152 | M1 | For matching velocities at
t = 15 and using v = 0 at t = 35
[1000b = -40 → b = –0.04]
[a = 0.04 × 352 = 49] | M1 | Solve for a and b
a = 49 and b = -0.04 AG | A1
Total: | 3
--- 5(ii) ---
5(ii) | 0 ⩽ t ⩽ 5 correct | B1 | Increasing quadratic, from (0,0) to
(5,20), concave up
5 ⩽ t ⩽ 15 correct | B1 | Line from (5,20) to (15,40)
15 ⩽ t ⩽ 35 correct | B1 | Decreasing quadratic, from (15,40) to
(35,0), concave down
20 and 40 seen correct on v-axis | B1
Total: | 4
--- 5(iii) ---
5(iii) | 5 100
A = ∫0.8t2dt =
1 3
0 | B1
A = 1( 20+40 )×10=300
2 2 | M1 | Using trapezium rule or integration for
t = 5 to t = 15
35( )
A = ∫ a+bt2 dt
3
15
0.04
=49t− t3
3 | M1 | Attempt to integrate the quadratic
function
from t = 15 to t = 35
A = 453.3333 = 1360/3
3 | A1
Total Distance = 2360/3 = 787m | A1
Total: | 5
Question | Answer | Marks | Guidance
A particle $P$ moves in a straight line starting from a point $O$ and comes to rest $35$ s later. At time $t$ s after leaving $O$, the velocity $v$ m s$^{-1}$ of $P$ is given by

$$v = \frac{4}{5}t^2 \quad 0 \leq t \leq 5,$$
$$v = 2t + 10 \quad 5 \leq t \leq 15,$$
$$v = a + bt^2 \quad 15 \leq t \leq 35,$$

where $a$ and $b$ are constants such that $a > 0$ and $b < 0$.

\begin{enumerate}[label=(\roman*)]
\item Show that the values of $a$ and $b$ are $49$ and $-0.04$ respectively. [3]

\item Sketch the velocity-time graph. [4]

\item Find the total distance travelled by $P$ during the $35$ s. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2017 Q5 [12]}}
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