Question 1 - A-Level Maths

OCR MEI M1 — Question 1 19 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Marks	19
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Travel graphs
Type	Velocity from displacement function using calculus
Difficulty	Moderate -0.3 This is a standard M1 kinematics question testing differentiation of polynomials to find velocity and acceleration, finding stationary points, and calculating displacement. All parts follow routine procedures with no novel problem-solving required, though part (vi) requires careful consideration of direction changes. Slightly easier than average due to straightforward calculus and algebraic manipulation.
Spec	3.02a Kinematics language: position, displacement, velocity, acceleration 3.02b Kinematic graphs: displacement-time and velocity-time 3.02c Interpret kinematic graphs: gradient and area

The displacement, $x$ m, from the origin O of a particle on the $x$-axis is given by $$x = 10 + 36t + 3t^2 - 2t^3,$$ where $t$ is the time in seconds and $-4 \leqslant t \leqslant 6$.

Write down the displacement of the particle when $t = 0$. [1]
Find an expression in terms of $t$ for the velocity, $v$ ms$^{-1}$, of the particle. [2]
Find an expression in terms of $t$ for the acceleration of the particle. [2]
Find the maximum value of $v$ in the interval $-4 \leqslant t \leqslant 6$. [3]
Show that $v = 0$ only when $t = -2$ and when $t = 3$. Find the values of $x$ at these times. [5]
Calculate the distance travelled by the particle from $t = 0$ to $t = 4$. [3]
Determine how many times the particle passes through O in the interval $-4 \leqslant t \leqslant 6$. [3]

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1(i)	0	B1
(ii)	v=36+6t−6t2	M1
A1	Attempt at differentiation	2
(iii)	a=6−12t	M1
F1	Attempt at differentiation	2
(iv)	Take a = 0

so t=0.5

Answer	Marks
and v = 37.5 so 37.5 m s -1	M1

A1

Answer	Marks
A1	Allow table if maximum indicated or implied

FT their a

cao Accept no justification given that this is

Answer	Marks
maximum	3
(v)	either

Solving 36+6t−6t2=0

so t = -2 or t = 3

or

Sub the values in the expression for

v

Both shown to be zero

A quadratic so the only roots

then

x(-2) = -34

Answer	Marks
x(3) = 91	M1

B1

E1

M1

E1

B1

Answer	Marks
B1	A method for two roots using their v

Factorization or formula or … of their expression

Shown

Allow just 1 substitution shown

Both shown

Must be a clear argument

cao

Answer	Marks
cao	5
(vi)	x(3)−x(0)+ x(4)−x(3)

= 91−10+74−91

Answer	Marks
= 98 so 98 m	M1

A1

Answer	Marks
A1	Considering two parts

Either correct

cao

Answer	Marks
[SC 1 for s(4) – s(0) = 64]	3

(vii)

Answer	Marks
P	At the SP of v

x(-2) = -34 i.e. < 0 and

x(3) = 91 i.e. > 0

Also x(-4) = 42 > 0 and

x(6) = -98 < 0

Answer	Marks
hsoy sthircese Atimneds MathsTutor.com	M1

B1

Answer	Marks
B1	Or any other valid argument e.g find all the zeros,

sketch, consider sign changes. Must have some

working. If only a sketch, must have correct shape.

Doing appropriate calculations e.g. find all 3 zeros;

sketch cubic reasonably (showing 3 roots); sign

changes in range

Answer	Marks
3 times seen	3

19

Answer	Marks
mark	Sub

Question 1:
--- 1(i) ---
1(i) | 0 | B1 | 1
(ii) | v=36+6t−6t2 | M1
A1 | Attempt at differentiation | 2
(iii) | a=6−12t | M1
F1 | Attempt at differentiation | 2
(iv) | Take a = 0
so t=0.5
and v = 37.5 so 37.5 m s -1 | M1
A1
A1 | Allow table if maximum indicated or implied
FT their a
cao Accept no justification given that this is
maximum | 3
(v) | either
Solving 36+6t−6t2=0
so t = -2 or t = 3
or
Sub the values in the expression for
v
Both shown to be zero
A quadratic so the only roots
then
x(-2) = -34
x(3) = 91 | M1
B1
E1
M1
E1
B1
B1
B1 | A method for two roots using their v
Factorization or formula or … of their expression
Shown
Allow just 1 substitution shown
Both shown
Must be a clear argument
cao
cao | 5
(vi) | x(3)−x(0)+ x(4)−x(3)
= 91−10+74−91
= 98 so 98 m | M1
A1
A1 | Considering two parts
Either correct
cao
[SC 1 for s(4) – s(0) = 64] | 3
(vii)
P | At the SP of v
x(-2) = -34 i.e. < 0 and
x(3) = 91 i.e. > 0
Also x(-4) = 42 > 0 and
x(6) = -98 < 0
hsoy sthircese Atimneds MathsTutor.com | M1
B1
B1 | Or any other valid argument e.g find all the zeros,
sketch, consider sign changes. Must have some
working. If only a sketch, must have correct shape.
Doing appropriate calculations e.g. find all 3 zeros;
sketch cubic reasonably (showing 3 roots); sign
changes in range
3 times seen | 3
19
mark | Sub

Show LaTeX source

The displacement, $x$ m, from the origin O of a particle on the $x$-axis is given by
$$x = 10 + 36t + 3t^2 - 2t^3,$$
where $t$ is the time in seconds and $-4 \leqslant t \leqslant 6$.

\begin{enumerate}[label=(\roman*)]
\item Write down the displacement of the particle when $t = 0$. [1]
\item Find an expression in terms of $t$ for the velocity, $v$ ms$^{-1}$, of the particle. [2]
\item Find an expression in terms of $t$ for the acceleration of the particle. [2]
\item Find the maximum value of $v$ in the interval $-4 \leqslant t \leqslant 6$. [3]
\item Show that $v = 0$ only when $t = -2$ and when $t = 3$. Find the values of $x$ at these times. [5]
\item Calculate the distance travelled by the particle from $t = 0$ to $t = 4$. [3]
\item Determine how many times the particle passes through O in the interval $-4 \leqslant t \leqslant 6$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q1 [19]}}

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