Notes

N.B. Accept column vectors throughout apart from in the answer to (c).

B1: cao.
N.B. Must be using correct vector notation, including brackets if using column vectors.

Notes

N.B. Accept column vectors throughout apart from in the answer to (c).

M1: For an unsimplified expression, using their \(\mathbf{r}\), with the square root

A1: Accept any surd equivalent isw
N.B. Must come from \(\mathbf{r} = (32\mathbf{i} - 16\mathbf{j})\)

Notes

N.B. Accept column vectors throughout apart from in the answer to (c).

M1: Both powers of \(t\) decreasing by 1 (but not just division by \(t\))
N.B. M0 if \(\mathbf{i}\) and/or \(\mathbf{j}\) are missing and never reappear.

A1: Must be in terms of \(t\), \(\mathbf{i}\) and \(\mathbf{j}\)

Notes

N.B. Accept column vectors throughout apart from in the answer to (c).

M1: Both powers of \(t\) decreasing by 1 (but not just division by \(t\))
If no i’s and/or j’s, can score M1 if
EITHER they have 2 separate components, provided they are clearly treated as such in the subsequent working
OR the i’s and j’s reappear
otherwise M0.

A1: Correct vector or
2 correct separate components, provided they are clearly treated as such in the subsequent working

M1: Complete method to form a correct equation in \(T\) (\(t\)) only, for their a, using the 2 components from their a

A1: Accept 20.3 or 20.25 or any equivalent fraction (allow \(t\) instead of \(T\))

Notes

Accept column vectors throughout

B1: \(2c\mathbf{i} - 6\mathbf{j}\) seen or implied.   B0 for \(\mathbf{r} = 2c - 6\) if no evidence of components.

M1: ALT 1: Use the bearing to obtain a CORRECT diagram showing a right-angled triangle with at least one \(45^\circ\) angle marked or clearly explained (i.e. \(135^\circ\) marked on the diagram and either \(135^\circ - 90^\circ = 45^\circ\) or \(180^\circ - 135^\circ = 45^\circ\)), and \(2c\) and \(\pm 6\) marked AND use of isosceles triangle or tan or (sin/cos and Pythag) to obtain \(2c = 6\)
ALT 2: No diagram required   Use \(\tan 135^\circ = \dfrac{2c}{-6} \Rightarrow 2c = 6\)

A1*: Given answer correctly obtained

ALTERNATIVE when \(t = 4\) is substituted at the end:

B1: \(ct^{\frac{1}{2}} = 2c\) and \((-)\dfrac{3t^2}{8} = (-)6\) when \(t = 4\), seen or implied

M1: ALT 3: Use the bearing to obtain a CORRECT diagram showing a right-angled triangle with at least one \(45^\circ\) angle marked or clearly explained (i.e. \(135^\circ\) marked on the diagram and either \(135^\circ - 90^\circ = 45^\circ\) or \(180^\circ - 135^\circ = 45^\circ\)), and \(ct^{\frac{1}{2}}\) and \(\pm\dfrac{3t^2}{8}\) marked AND use of isosceles triangle or tan or (sin/cos and Pythag) to obtain \(2c = 6\) when \(t = 4\)

A1*: Given answer correctly obtained

Notes

Accept column vectors throughout

M1: Both powers of \(t\) decreasing by 1 (M0 if \(\mathbf{i}\) or \(\mathbf{j}\) is missing but allow recovery or working with components only)
N.B. This mark is available if \(c\) has not been substituted for.

A1: Correct unsimplified derivative or two correct components

M1: Put \(t = 4\) in their \(\mathbf{v}\) (must be using an attempted derivative of \(\mathbf{r}\)) and then use Pythagoras with the root, allow a missing \(-\) sign
N.B. If they state \(t = 4\), allow a slip when they substitute in, for this M mark.
This mark is available if \(c\) has not been substituted for.

A1: Accept 3.1 or better

Notes

Accept column vectors throughout

M1: Both powers of \(t\) decreasing by 1
N.B. This mark is available if \(c\) has not been substituted for
(M0 if \(\mathbf{i}\) or \(\mathbf{j}\) is missing but allow recovery or working with components only).

A1: Correct unsimplified derivative

M1: Use of an appropriate ratio (must be using an attempted derivative of their \(\mathbf{v}\)), condone sign error and the reciprocal, to obtain an equation in \(t\) or \(T\) only.
N.B. If they state that \(-\dfrac{3}{4}T^{-\frac{3}{2}}\mathbf{i} - \dfrac{3}{4}\mathbf{j} = k(-\mathbf{i} - 27\mathbf{j})\) and then equate coefficients to give two simultaneous equations in \(k\) and \(T\), these need to be used to produce an equation in \(T\) only, before the M mark is earned.

A1: cao (allow \(t\) instead of \(T\))

Notes

Allow column vectors throughout.

B1: cao

M1: Use of Pythagoras, including the square root, on a velocity vector at \(t = 0\)

A1: cao. Must come from a correct \(\mathbf{v}\).

Notes

Allow column vectors throughout.

M1: Equating \(\mathbf{i}\) and \(\mathbf{j}\) components of \(\mathbf{v}\) or a ratio of 1:1 to obtain a quadratic in \(t\) only.
If they use a constant, e.g. \(t^2 - 3t + 7 = k\) and \(2t^2 - 3 = k\), \(k\) must be eliminated to earn this mark.
N.B. M0 (since wrong working seen) if they write down
\(\mathbf{i} + \mathbf{j} = (t^2 - 3t + 7)\mathbf{i} + (2t^2 - 3)\mathbf{j}\)
OR \(\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}t^2 - 3t + 7\\2t^2 - 3\end{pmatrix}\)
OR \(t^2 - 3t + 7 = 1\) and \(2t^2 - 3 = 1\)
and then \(t^2 - 3t + 7 = 2t^2 - 3\)

A1: \(t = 2\)

N.B. Allow M1A1 for a correct trial and error method where they obtain \(\mathbf{v} = 5\mathbf{i} + 5\mathbf{j}\) when \(t = 2\) but M0 if they don’t get \(t = 2\)

Notes

Allow column vectors throughout.

M1: At least one power decreasing by 1 in each component in their \(\mathbf{v}\)
(M0 if clearly dividing by \(t\))
Both \(\mathbf{i}\) and \(\mathbf{j}\) needed in their answer or a column vector
Allow recovery if the \(\mathbf{i}\) and \(\mathbf{j}\) disappear and then reappear.

A1: cao (must be a vector) isw e.g. if they find the magnitude or put \(t = 0\) or differentiate again
\(\mathbf{i}\)’s and \(\mathbf{j}\)’s do not need to be collected.
N.B. Allow M1A0 for \(2t - 3\mathbf{i} + 4t\mathbf{j}\)

Notes

Allow column vectors throughout.

M1: \(2t - 3 = 0\) or (their derivative of the \(\mathbf{i}\)-component of \(\mathbf{v}\)) = 0
N.B. M0 if they equate the derivative of both components of \(\mathbf{v}\) to zero.

A1: cao
N.B. Correct answer, with no working, can score both marks.

Notes

Accept column vectors throughout apart from the answer to (b).

M1: Need square root but -ve sign not required. Allow \(\mathbf{i}\)’s and/or \(\mathbf{j}\)’s to go missing from their \(\mathbf{v}\) at \(t = 2\), provided they have applied Pythagoras correctly.

A1: cao
N.B. Correct answer with no working can score 2 marks.

Notes

Accept column vectors throughout apart from the answer to (b).

M1: Both powers decreasing by 1. Allow a column vector.
M0 if \(\mathbf{i}\) or \(\mathbf{j}\) is missing but allow recovery in (b).

A1: cao. Do not accept a column vector.

Notes

Accept column vectors throughout apart from the answer to (b).

M1: Both powers increasing by 1
M0 if \(\mathbf{i}\) or \(\mathbf{j}\) is missing but allow recovery.

A1: (\(\mathbf{r} =\)) not required

M1: Putting \(\mathbf{r} = (\mathbf{i} - 4\mathbf{j})\) and \(t = 4\) into their displacement vector expression which must have \(\mathbf{C}\) (allow \(C\)) to give an equation in \(\mathbf{C}\) only, seen or implied.
Must have attempted to integrate \(\mathbf{v}\) for this mark to be available.
N.B. \(\mathbf{C}\) does not need to be found and this is a method mark, so allow slips.

A1: cao

Notes

Allow column vectors throughout this question

M1: Both powers decreasing by 1 (M0 if vector(s) disappear but allow recovery)

A1: cao

Notes

Allow column vectors throughout this question

M1: Complete method, using \(\mathbf{v}\), to obtain an equation in \(t\) only, allow a sign error

DM1: Dependent on M1, solve for \(t\)

A1: cao

Notes

Allow column vectors throughout this question

M1: Both powers increasing by 1 (M0 if vectors disappear but allow recovery)

A1: Correct expression without \(\mathbf{C}\)

A1: cao

Notes

Allow column vectors throughout this question

M1: Use of Pythagoras on \(\mathbf{v}\) and 10 to set up equation in \(t\)

M(A)1: Correct 3 term quadratic in \(t\)

A1: cao

M1: Substitute their numerical \(t\) value into their \(\mathbf{r}\)

M1: Use of Pythagoras to find the magnitude of their \(\mathbf{r}\)

A1: cso

Notes

Accept column vectors throughout

M1: At least 3 terms with powers increasing by 1 (but M0 if clearly just multiplying by \(t\))

A1: Correct expression

A1: Accept \(8\mathbf{i} - 9.3\mathbf{j}\) or better. Isw if speed found.

Notes

Accept column vectors throughout

M1: Must have an equation in \(t\) only (Must have integrated to find a velocity vector)

A1ft: Correct equation follow through on their \(\mathbf{v}\) but must be a 3 term quadratic

A1: cao

Notes

Accept column vectors throughout

M1: At least 2 terms with powers decreasing by 1 (but M0 if clearly just dividing by \(t\))

A1: Correct expression

M1: Use magnitude to give an equation in \(t\) only, must have differentiated to find a velocity (M0 if they use \(\sqrt{x^2 - y^2}\))

A1: Correct equation \(\sqrt{(2t - 1)^2 + 3^2} = 5\)

M1: Solve a 3 term quadratic for \(t\) which has come from differentiating and using a magnitude. This M mark can be implied by a correct answer with no working.

A1: 2.5

Notes

N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers

M1: Use of \(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\) with attempt to differentiate (both powers decreasing by 1)
M0 if \(\mathbf{i}\)’s and \(\mathbf{j}\)’s omitted and they don’t recover

A1: Correct differentiation in any form

A1: Correct and simplified.
Ignore subsequent working (ISW) if they go on and find the magnitude.

Notes

N.B. Accept column vectors throughout and condone missing brackets in working but they must be there in final answers

M1: Use of \(\mathbf{r} = \displaystyle\int \mathbf{v}\,\mathrm{d}t\) with attempt to integrate (both powers increasing by 1)
M0 if \(\mathbf{i}\)’s and \(\mathbf{j}\)’s omitted and they don’t recover

A1: Correct integration in any form. Condone \(\mathbf{r}_0\) not present

A1: Correct and simplified.

Typical solution

Typical solution

When \(t = 0,\ \mathbf{v} = 4\mathbf{j},\ c_2 = 4\)

When \(t = 2\),

Typical solution

Using \(a = \dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}\)

Using \(a = -1.8\) when \(t = 3\)

Typical solution

Typical solution

\(-8\)

Typical solution

So \(0 \lt t \lt 3\)

Typical solution

\(v = 1\) when \(t = 0\) then \(c = 1\)

\(v\) = 10 and \(t\) = 3

Typical solution

Notes

B1: Correct expression for the acceleration of the particle

B0 if \(s = 3 + 1.2t\)

M1*: Sets their linear expression for \(a\) equal to 15

For reference \(t = 10\)

M1*: Attempt to integrate expression – at least three terms with correct \(t^2\) or \(t^3\) term

A1: Correct expression for the displacement (allow un-simplified)

A0 if \(a = \int v\,\mathrm{d}t\)

M1dep*: Correct use of limits with their positive value of \(t\) – must see as a minimum all three values (either simplified or un-simplified), for example, if correct then simplified is \(20 + 150 + 200\). No consideration of lower limit of zero is required but if considered then must be correct, for example, \(2(10) + \frac{3}{2}(10)^2 + \frac{0.6}{3}(10)^3 - 2(0) + \frac{3}{2}(0)^2 + \frac{0.6}{3}(0)^3\) (so incorrect bracketing) is M0 (as DR)
M0 for \(20 + 150 + 200 + c\) unless \(c = 0\) is explicitly stated (so not just implied by seeing 370 as a final answer)

A correct answer of 370 does not imply this mark
This M mark is dependent on both previous M marks

B1: No justification that \(v \gt 0\) for all values of \(t\) is required for full marks. Ignore any working seen for this mark but full marks can only be awarded for a completely correct solution

Notes

No scaling on axes required for the first two B marks (so ignore any scaling for the first two marks)

B1: Quadratic curve (positive quadratic) passing through the origin and appearing in both quadrants with a minimum turning point in the fourth quadrant

B1: Inverse proportional (to \(t^2\)) curve (so correct curvature), should not touch the horizontal axis
Condone if it appears to be approaching a horizontal asymptote (other than the \(t\)-axis) for this mark but gradient must not be positive

B1: Completely correct graph with both curves meeting at 6 with values of 3 and 6 on \(t\) axis (or clearly stated in their working in part (a) only) – this mark is dependent on the previous 2 B marks
Allow only the \(t\)-axis as a horizontal asymptote
Value of 9 not required on \(t\) axis, neither is \(\frac{3}{2}k\) on the \(v\)-axis

Notes

B1: From symmetry or solving \(\frac{1}{12}k(2t - 3) = 0\) www
Any other answer(s) is B0

Notes

M1*: Attempt to integrate – both terms with power increased by 1 with one term correct
Ignore \(+c\) omission for M marks

B1: www
Condone \(\displaystyle [k]\tfrac{1}{12}\int_0^3 \left(t^2 - 3t\right)\mathrm{d}t = +\frac{3}{8}[k]\)
Award B1 for \(\displaystyle [k]\tfrac{1}{12}\int_0^6 \left(t^2 - 3t\right)\mathrm{d}t = \frac{18}{8}[k]\) but B0 for \(\displaystyle [k]\tfrac{1}{12}\int_0^6 \left(t^2 - 3t\right)\mathrm{d}t = \frac{12}{8}[k]\)
The correct values do not imply the M mark as DR

M1*: Attempt to integrate – answer of the form \(ct^{-1}\) with \(c \ne 1, 54k, 54\)

B1: www
The correct value does not imply the M mark as DR

M1dep*: Forming a linear equation in \(k\) with the correct number of relevant terms (e.g. must have taken the modulus of their integral from 0 to 3)
\(\displaystyle \tfrac{1}{12}k\int_0^6 \left(t^2 - 3t\right)\mathrm{d}t = \frac{12}{8}k\), or any working that suggests the modulus was not taken between 0 and 3 is M0

B1: www

Additional guidance for 12(c) (Appendix)

The square brackets around the \(k\) in the main scheme means that this may be omitted as they could deal with this constant at the end when they form their equation with the 84 (as the \(k\) is a constant factor for both expressions for \(v\))

The first B mark is for a correct expression/value for both \(\displaystyle I_1 = [k]\tfrac{1}{12}\int_0^3 \left(t^2 - 3t\right)\mathrm{d}t\) and \(\displaystyle I_2 = [k]\tfrac{1}{12}\int_3^6 \left(t^2 - 3t\right)\mathrm{d}t\) in any un-simplified form:
so for \(I_1\) any expression that is equivalent to \(\pm\frac{1}{12}\left(\frac{3^3}{3} - \frac{3(3)^2}{2}\right)[k]\) is fine (allow either positive or negative due to the fact that this area is below the \(t\)-axis) e.g. \(I_1 = \pm\frac{1}{12}\left(9 - \frac{27}{2}\right)[k]\) or \(\pm\frac{9}{24}[k]\) etc. (ISW once a correct un-simplified form is seen)

For \(I_2\) any expression that is equivalent to \(\frac{1}{12}\left(\frac{6^3}{3} - \frac{3(6)^2}{2}\right)[k] - \frac{1}{12}\left(\frac{3^3}{3} - \frac{3(3)^2}{2}\right)[k]\) is fine e.g. \(\frac{1}{12}(72 - 54)[k] - \frac{1}{12}\left(9 - \frac{27}{2}\right)[k]\) or \(\left(\frac{3}{2} + \frac{3}{8}\right)[k]\) etc.
If the candidate does not consider these two integrals separately and instead attempts to combine as a single integral (between 0 and 6) then they must consider it correctly (given the applied context) so e.g.
\(\displaystyle I_{1,2} = [k]\tfrac{1}{12}\int_0^6 \left(t^2 - 3t\right)\mathrm{d}t = -\tfrac{1}{12}\left(\tfrac{3^3}{3} - \tfrac{3(3)^2}{2}\right)[k] + \tfrac{1}{12}\left(\tfrac{6^3}{3} - \tfrac{3(6)^2}{2}\right)[k] - \tfrac{1}{12}\left(\tfrac{3^3}{3} - \tfrac{3(3)^2}{2}\right)[k]\) scores B1 but \(+\tfrac{1}{12}\left(\tfrac{3^3}{3} - \tfrac{3(3)^2}{2}\right)[k] + \tfrac{1}{12}\left(\tfrac{6^3}{3} - \tfrac{3(6)^2}{2}\right)[k] - \tfrac{1}{12}\left(\tfrac{3^3}{3} - \tfrac{3(3)^2}{2}\right)[k]\) is B0

The second B mark is for a correct expression/value for \(\displaystyle I_3 = 54[k]\int_6^9 t^{-2}\,\mathrm{d}t\) in any un-simplified form: e.g. \(I_3 = 54\left(-\frac{1}{9} - \left(-\frac{1}{6}\right)\right)[k]\) or \(54\left(-\frac{1}{9} + \frac{1}{6}\right)[k]\) etc. (ISW once a correct un-simplified form is seen)

The third M mark is for considering \(|I_1| + I_2 + I_3 = 84\) to form a linear equation in \(k\) (with the correct number of relevant terms) – this mark is dependent on the first two M marks and also they must have taken the modulus or equivalent for their integral between 0 and 3 for this mark e.g. they must have considered \(\displaystyle -[k]\tfrac{1}{12}\int_0^3 \left(t^2 - 3t\right)\mathrm{d}t\) or \(\displaystyle [k]\tfrac{1}{12}\int_3^0 \left(t^2 - 3t\right)\mathrm{d}t\) oe

As this question is detailed reasoning the stages as shown in the MS must all be done to award each corresponding mark. So, an answer of \(k = 16\) with no working scores B1 only.

Notes

M1: Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(a = \pm 10\) or \(\pm 9.8\) or \(\pm g\), and \(u = 39\sin\theta\) or \(u = 39\cos\theta\).
Accept any other complete method (using correct suvat equations) to find the maximum height e.g. \(0 = 39\sin\theta - 10t\) and with \(t\) then substituted into \(s = (39\sin\theta)t + 0.5(-10)t^2\)
Condone \(g = 9.8\) for full marks in (a) and (b)

A1: A correct equation with the correct value of sin substituted or for \(0 = (39\sin(22.6\ldots))^2 + 2(-10)h\)
Allow using \(g\) (and not replaced with 10 or 9.8) for this A mark

A1: www

• accept awrt 31.2 (using \(22.6\ldots\) as the angle and \(g = 10\))
• accept awrt 31.5 (coming from exact value of sin or \(22.6\ldots\) and \(g = 9.8\))
• condone 31.3 (3 sf)

Notes

M1: Applying \(s = ut + \frac{1}{2}at^2\) with \(s = \pm 20\), \(a = \pm 10\) or \(\pm 9.8\) or \(\pm g\) and \(u = 39\sin\theta\) or \(u = 39\cos\theta\)
Condone \(g = 9.8\) for full marks in (a) and (b)
Accept any other complete method to find \(T\)

A1: With correct value of sine or \(\sin(22.6)\)
Allow \(-20 = \left(39 \times \frac{5}{13}\right)T + \frac{1}{2}(-9.8)T^2\)
Allow using \(g\) (and not replaced with 10 or 9.8) for this A mark

A1: BC

• accept awrt 4.07 (using \(g = 9.8\) and exact value of sine)
• accept awrt 4.06 (using \(g = 9.8\) and \(22.6\ldots\))
• a value of \(4(.00\ldots)\) coming from \(3.998\ldots\) (using \(g = 10\) and \(22.6\ldots\) for sine)

Condone \(t = 4\)

Notes

B1: Allow any correct limitation
B0 if referring to

• Air resistance and/or wind (only)
• The ground is unlikely to be horizontal (only)
• The mass or weight or shape of \(P\) (only)
• The angle/heights/speeds may not be as quoted (only)
• Modelling the problem as 3D rather than in 2D (only)

If multiple limitations given, and any are incorrect, then B0

Notes

M1: Differentiate given \(v\) (at least two terms correct)

M1*: Applying \(s = ut\) horizontally to find distance \(BC\) with correct value of cos (or \(\cos(22.6\ldots)\)) and their value of \(T\) from (b)
Allow \(g = 9.8\) which if correct leads to \(146.34\ldots\)

M1*: Integrate given \(v\) (at least two terms correct)

M1dep*: Puts their integrated expression for \(s\), with \(t =\) their \(T\) from (b), equal to their distance for \(BC\) to form an equation in \(k\) only – dependent on the two previous M marks only
Must not include a \(+c\) unless dealt with as part of a definite integral. However, if a \(+c\) is included then subsequently ignored/set equal to zero without justification then give bod for this and any subsequent A marks (if earned)

A1: Correct exact value for \(k\) (oe e.g. 0.0625)
Final two marks can only be awarded if \(g = 10\) used

A1: oe

Notes

M1: Considers either the i or j component equal to zero or forms a five-term quartic equation for \(|\mathbf{v}|^2 = 0\) (oe)
\((4t^4 + 4t^3 - 47t^2 - 30t + 170 = 0)\)

A1: BC – need not see the negative value of \(t\) or for substituting \(t = 0.5\) into quadratic expression for \(\mathbf{j}\) and showing this gives a non-zero answer (oe) with correct working and conclusion
A1 for the correct quartic equation with roots stated as \(2.3 \pm 0.35\mathrm{i}\), \(-2.8 \pm 0.65\mathrm{i}\) + correct conclusion

Notes

B1: Correct derivative – or as a column vector
Brackets must be around the \(4t + 1\)

Notes

M1*: Setting up a quadratic equation in \(t\) only – allow sign errors (including on the 1 and 2) and the 1 and \(-2\) on the wrong side
Or multiples of 1 and \(-2\)

M1dep*: Solves their (two or three term) quadratic and selects their positive value of \(t\)
Check unsupported solutions if incorrect quadratic equation

M1*: Substitute their \(\mathbf{a}\) into \(\mathbf{F} = 0.5\mathbf{a}\) or their \(|\mathbf{a}|\) into \(|\mathbf{F}| = 0.5|\mathbf{a}|\). If \(\mathbf{F}\) not stated in terms of \(t\) then one component must be correct following through from their \(\mathbf{a}\) (and possibly \(t\))
Must use correct value of 0.5 for \(m\) but can be in terms of \(t\)

M1dep*: Dependent on previous M mark only
From a value of \(t \gt 0\)

A1: awrt 5.59 (exact: \(\dfrac{5\sqrt{5}}{2}\))
5.590169…

Notes

M1*: Integrates \(\mathbf{v}\) wrt \(t\) – at least three terms correct
Allow without \(+\mathbf{c}\)

A1: Uses given conditions to find correct \(\mathbf{c}\) – dependent on a completely correct integrated expression for \(\mathbf{s}\)
www

M1dep*: Substitute \(t = 1.5\) into their \(\mathbf{s}\)

M1: Attempt to find a relevant angle using the components of their \(\mathbf{s}\) (allow use of sin/cos with the magnitude of \(\mathbf{s}\))
Dependent on both previous M marks.
Written in terms of arctan is sufficient

A1: awrt 190 (or from \(270 - \tan^{-1}\left(\frac{33/8}{3/4}\right)\))
190.3048465…

Notes

M1: Combines given forces and considers either component equal to zero – allow M1 for either \(4t - 8 = 0\) or for \(6(2t - 1)^2 - 54 = 0\)

A1: \(t = 2\) only – need only consider \(\mathbf{i}\) or \(\mathbf{j}\) component but any contradictory working/answers scores A0

Notes

B1: Using \(\mathbf{F} = 2\mathbf{a}\) correctly
Allow \(2\mathbf{a} = \ldots\)

M1*: Attempt to integrate \(\mathbf{a}\) (or \(\mathbf{F}\)) wrt \(t\) – two of their terms integrated correctly
M0 if only considering one force or one component for \(\mathbf{a}\) or \(\mathbf{F}\)

A1: Condone no \(+\mathbf{c}\) for this mark
\(\mathbf{v} = \left(t^2 - 4t\right)\mathbf{i} + \left(4t^3 - 6t^2 - 24t\right)\mathbf{j}\) (oe)
Allow \(2\mathbf{v} = \ldots\)

M1dep*: Uses correct initial conditions to find \(\mathbf{c}\) (or \(\mathbf{c} = \mathbf{0}\) if expanded version used)

M1: Sets \(\mathbf{i}\)-component of \(\mathbf{v}\) equal to 0 to obtain a quadratic equation in \(t\)
Dependent on first M mark

A1: Must have found \(+\mathbf{c}\) for this mark

Notes

M1*: Attempt to integrate \(\mathbf{v}\) wrt \(t\) – two of their terms integrated correctly – dependent on first M mark in (b)
no vector constant of integration required in (c)

A1: \(\mathbf{r} = \left(\frac{1}{3}t^3 - 2t^2\right)\mathbf{i} + \left(t^4 - 2t^3 - 12t^2\right)\mathbf{j}\)

M1dep*: Attempt to find \(\mathbf{r}\) at both \(t = 0\) and \(t = 3\)

M1: Correct expression for the distance between given times (dependent on both previous M marks)

A1: (For reference: 81.49846624…)
\(\sqrt{6642} = 9\sqrt{82}\)

Notes

M1: Finds displacement as vector or in two directions separately
Soi
Allow use of \(\mathbf{i}\) and \(\mathbf{j}\) vectors throughout

M1: Pythagoras’ used

A1: Must be scalar answer

Notes

M1: Attempt to differentiate at least one component
soi

A1: One correct component or scalar in one direction

A1: Both correct components in a vector

Notes

B1: Either value seen

B1: Allow for vector or both components clear which is horizontal and which vertical
Do not allow if the final answer is scalar

Notes

B1: Must indicate the resistance has a horizontal component.
Do not accept “air resistance” on its own.

SC1 “Factor \(e^{-kt}\) should be included” or similar

Notes

M1: Attempt to integrate velocity either as a vector or 2 separate components

A1: Condone missing constant

M1: Either as a vector constant or 2 separate components evaluated.
May be implied by correct vector answer

A1: Must be in vector form (could be column vector but must be exact vector notation eg brackets and not \(\mathbf{i}\) and \(\mathbf{j}\) as well)
Allow \(3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + 7\mathbf{j}\)

Notes

M1: Attempt to eliminate \(t\) from the parametric equations

A1: FT their (a)
Isw

Notes

M1: differentiating the \(\mathbf{v}\) vector

M1: substituting \(t = 2\) into their \(\mathbf{a}\) vector

M1: Equate the magnitude of their \(\mathbf{a}\) vector to 10

A1: must choose the positive value if two values seen

Notes

M1: integrating with their \(k\) or general \(k\). Allow for a vector or for both components separately integrated.

A1: Condone missing \(+\mathbf{c}\) or \(+\mathbf{c}\) still in their answer
FT their \(k\) if positive or general \(k\) used
Must be in vector form

Notes

M1: FT their \(\mathbf{r}\)

A1: www

Notes

M1*: Attempt to integrate both components; condone missing \(+\mathbf{c}\)

M1 dep*: Using initial conditions

M1 dep*: Using \(t = 2\) to find position vector or values for \(x\) and \(y\)

A1: Accept vector form or two clear components.

M1: Using Pythagoras

A1: FT their components

Notes

M1: Extracting equations for \(x\) and \(y\) from their displacement vector

M1 dep: Attempt to eliminate \(t\)

A1: cao

Equivalent form \(y = 7(x+1)^{\frac{1}{3}} + 2\) for M1M1A0

Notes

M1*: Must be vector acceleration

M1*: Evaluating when \(t = 2\)
\(a = 12\) is sufficient here

M1*: Newton’s second law in vector form, or in \(x\)-direction only
If their \(\mathbf{a}\) has two non-zero components, allow for dividing 48 by the magnitude of their \(\mathbf{a}\)

A1 dep*: cao

Notes

M1: Attempt to integrate, condone omission of +c

M1: Attempt to evaluate \(c\)

A1: Any form

M1: Forming an equation using their \(v = 0\)

M1: Use of discriminant or completing the square, showing equation has complex roots or stating that the equation has no real roots
Allow this M mark for solving their equation if it has real solutions

E1: Clear conclusion in context consistent with their working.
FT their \(v\). Dependent on at least 1 method mark.
Ignore any reference to \(t < 0\) but do not allow stationary at \(t = 0\)

(corrected from the printed mark scheme: the discriminant is printed as \(0.5^5 - 4 \times 0.4 \times 3\); it should be \(0.5^2\))

Alternative (OR)

M1: Attempt to integrate, condone omission of +c

M1: Attempt to evaluate \(c\)

M1: Uses the positivity of \(t\) to establish the positivity of \(v\).

M1: Argues that \(v\) is always positive

E1: Clear conclusion in context consistent with their working.
FT their \(v\). Dependent on at least 1 method mark.

Alternative (OR)

M1: Attempt to construct an argument based on the positivity of \(v\).

M1: Uses the positivity of \(t\) aiming to establish the positivity of \(a\)

A1: Clear argument that \(a > 0\)

M1: Uses the link between \(a > 0\) and \(v\)

M1: Uses \(v_0 = 3\) explicitly in their argument

E1: Convincing complete argument.