Question 7 - A-Level Maths

CAIE M1 2019 June — Question 7 11 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2019
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Travel graphs
Type	Two-particle meeting or overtaking
Difficulty	Moderate -0.3 This is a standard velocity-time graph question requiring sketching linear functions, integrating to find displacement, and finding maximum separation. While it involves multiple steps (11 marks total), each component uses routine A-level mechanics techniques with no novel insight required—slightly easier than average due to straightforward linear velocities and clear structure.
Spec	1.08d Evaluate definite integrals: between limits 3.02a Kinematics language: position, displacement, velocity, acceleration 3.02b Kinematic graphs: displacement-time and velocity-time

Particles \(P\) and \(Q\) leave a fixed point \(A\) at the same time and travel in the same straight line. The velocity of \(P\) after \(t\) seconds is \(6(t - 3)\) m s\(^{-1}\) and the velocity of \(Q\) after \(t\) seconds is \((10 - 2t)\) m s\(^{-1}\).

Sketch, on the same axes, velocity-time graphs for \(P\) and \(Q\) for \(0 \leq t \leq 5\). [3]
Verify that \(P\) and \(Q\) meet after 5 seconds. [4]
Find the greatest distance between \(P\) and \(Q\) for \(0 \leq t \leq 5\). [4]

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks
7(i)	Straight line, reaching positive v-axis and positive t-axis (negative
gradient)	B1
Quadratic (U shape, through (0,0) and cutting t-axis at t < 5)	B1

Fully correct graphs with correct labelling with

Answer	Marks
t = 3, t = 5, v = 10, v = 60 seen	B1

3

Answer	Marks	Guidance
7(ii)	s=∫ ( 10−2t ) dt =10t−t2 (+ c)
or use area of a triangle ½ × 10 × 5 [= 25]	B1	Use either integration to find s for Q or use a correct formula

to find the area under the relevant triangle

Answer	Marks
M1	Use integration to find the displacement for P

( )

Answer	Marks	Guidance
s=∫ 6t2 −18t dt =2t3 −9t2 (+c )	A1	Correct integration for P (unsimplified)

s ( P )=2t3−9t2 5 =25

 

0 or solve

Answer	Marks	Guidance
10t−t2 =2t3−9t2	B1	Either

evaluation of s(P) at t = 5 and show that at t = 5, s(P) = s(Q)

= 25

or show that t = 5 is a solution of the cubic by solving

or verify t = 5 is a solution of the cubic by substitution.

4

Answer	Marks	Guidance
Question	Answer	Marks

Answer	Marks	Guidance
7(iii)	Distance PQ =	s – s
P Q	M1	Find the distance between P and Q Allow either sign

s and s must have been found by integration

P Q

Answer	Marks	Guidance
Maximum s if 6t 2 – 16t – 10 = 0	M1	Differentiate to obtain an equation in t and attempt to solve
t = 3.19	A1
Maximum Distance PQ = (–)48.4 m	A1

Alternative method for question 7(iii)

Answer	Marks	Guidance
6t2 – 18t = 10 – 2t	M1	State that greatest distance between P and Q occurs when v

P

= v

Q

Answer	Marks	Guidance
6t2 – 16t – 10 = 0	M1	Rearrange and attempt to solve for t
t = 3.19	A1
Maximum Distance PQ = (–)48.4 m	A1

4

Question 7:
--- 7(i) ---
7(i) | Straight line, reaching positive v-axis and positive t-axis (negative
gradient) | B1
Quadratic (U shape, through (0,0) and cutting t-axis at t < 5) | B1
Fully correct graphs with correct labelling with
t = 3, t = 5, v = 10, v = 60 seen | B1
3
--- 7(ii) ---
7(ii) | s=∫ ( 10−2t ) dt =10t−t2 (+ c)
or use area of a triangle ½ × 10 × 5 [= 25] | B1 | Use either integration to find s for Q or use a correct formula
to find the area under the relevant triangle
M1 | Use integration to find the displacement for P
( )
s=∫ 6t2 −18t dt =2t3 −9t2 (+c ) | A1 | Correct integration for P (unsimplified)
s ( P )=2t3−9t2 5 =25
 
0
or solve
10t−t2 =2t3−9t2 | B1 | Either
evaluation of s(P) at t = 5 and show that at t = 5, s(P) = s(Q)
= 25
or show that t = 5 is a solution of the cubic by solving
or verify t = 5 is a solution of the cubic by substitution.
4
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | Distance PQ = |s – s | = ±(2t 3 – 8t 2 – 10t)
P Q | M1 | Find the distance between P and Q Allow either sign
s and s must have been found by integration
P Q
Maximum s if 6t 2 – 16t – 10 = 0 | M1 | Differentiate to obtain an equation in t and attempt to solve
t = 3.19 | A1
Maximum Distance PQ = (–)48.4 m | A1
Alternative method for question 7(iii)
6t2 – 18t = 10 – 2t | M1 | State that greatest distance between P and Q occurs when v
P
= v
Q
6t2 – 16t – 10 = 0 | M1 | Rearrange and attempt to solve for t
t = 3.19 | A1
Maximum Distance PQ = (–)48.4 m | A1
4

Show LaTeX source

Particles $P$ and $Q$ leave a fixed point $A$ at the same time and travel in the same straight line. The velocity of $P$ after $t$ seconds is $6(t - 3)$ m s$^{-1}$ and the velocity of $Q$ after $t$ seconds is $(10 - 2t)$ m s$^{-1}$.

\begin{enumerate}[label=(\roman*)]
\item Sketch, on the same axes, velocity-time graphs for $P$ and $Q$ for $0 \leq t \leq 5$. [3]
\item Verify that $P$ and $Q$ meet after 5 seconds. [4]
\item Find the greatest distance between $P$ and $Q$ for $0 \leq t \leq 5$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2019 Q7 [11]}}

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