M7.4 Calculus in kinematics

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (a), the majority of candidates were able to differentiate s correctly and equate it to zero to produce a quadratic.However, a significant proportion of students did not show any method to justify their solutions from their quadratic, ignoring the instruction at the top of the question and lost two marks. A significant number of candidates did not find a derivative, choosing instead to equate s to zero. This led to t values of 0 and 3 which then had a knock on effect on the marks that could be scored in part (b).

Common mistakes

• v = ds/dt = 6t² − 24t + 18 = 0, t = 1 or t = 3
• working (e.g. factorising) must be shown, not just calculator roots
• setting s = 0 instead of v = 0 (giving t = 0 and 3)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. There was the occasional candidate who incorrectly differentiated, missing out the “+18” term from their answers. This also impacted the marks that could be scored for part (b). Part (b) was challenging and candidates who sketched a graph of the function and considered areas were usually the most successful. The most common error was to just use 0s and 4s while a number integrated s. Some who had made progress used 8 - for the middle section, forgetting that the distance was required. A significant number made no attempt at this part.

Common mistakes

• P changes direction at t = 1 and t = 3: s₁ = 8, s₃ = 0, s₄ = 8, so total distance = 8 + 8 + 8 = 24 m
• most used only s₀ and s₄ (displacement)
• integrating s
• using −8 for the middle section
• a sketch helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (c), many found the acceleration but a number then failed to equate it to zero and so did not achieve any marks. Some did obtain t = 2 successfully but then did nothing with it. A few used the symmetry or completed the square and got to the required answer very quickly. The odd negative answer for speed was seen.

Common mistakes

• maximum speed where a = dv/dt = 12t − 24 = 0 (or by symmetry or completing the square v = 6(t − 2)² − 6): t = 2, speed 6 m s⁻¹ (must be positive)
• many found a but did not set it to 0, or stopped at t = 2

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

exact (surd) answer was required so those who only stated a decimal approximation did not sqrt 322 -162 thereby not gaining both marks. gain the accuracy mark. A few did not account properly for the negative sign and evaluated In the third part the majority of learners recognised the requirement to differentiate to find the velocity vector and most did so successfully. Occasionally i and j were omitted entirely resulting in an expression which was no longer a vector and the loss of the method mark. Attempts at suvat or integration were seen but only rarely. Virtually all gave their answer in terms of i, j and t as required. In part (d) those learners who achieved the marks in part (c) generally achieved the first two 2 was not here for differentiating to find the acceleration vector. Occasionally the power of 1 handled correctly but differentiation skills were mostly sound. The final two marks proved more challenging and many either could not relate the gradient to their coefficients in i and j or simply stopped after finding the acceleration vector. For those who did attempt a solution, some recognised the need to use the ratio of their i and j coefficients but the majority did not Other errors seen included ignoring i and j and setting the expression equal to ±3 or ± 1 identify and use the correct perpendicular gradient. Some just equated i and j coefficients. 3. Of those who achieved the correct equation in T, a few solved it incorrectly as a result of errors 1 -. A very small number of learners used the scalar product successfully. in handling T 2

Common mistakes

• differentiate r: v = 6t^(1/2) i − 2t j
• i and j must stay in the answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (d) those learners who achieved the marks in part (c) generally achieved the first two 2 was not here for differentiating to find the acceleration vector. Occasionally the power of 1 handled correctly but differentiation skills were mostly sound. The final two marks proved more challenging and many either could not relate the gradient to their coefficients in i and j or simply stopped after finding the acceleration vector. For those who did attempt a solution, some recognised the need to use the ratio of their i and j coefficients but the majority did not Other errors seen included ignoring i and j and setting the expression equal to ±3 or ± 1 identify and use the correct perpendicular gradient. Some just equated i and j coefficients. 3. Of those who achieved the correct equation in T, a few solved it incorrectly as a result of errors 1 -. A very small number of learners used the scalar product successfully. in handling T 2

Common mistakes

• a = 3t^(−1/2) i − 2j
• direction perpendicular to y = x/3 has gradient −3, so (j component)/(i component) = −3
• wrong: equating i and j coefficients or setting the expression to ±3 or ±1/3
• handling T^(−1/2)
• T = 81/4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), most differentiated the given expression for v with respect to t to get to an expression for their acceleration. Candidates usually handled the powers correctly but there were the occasional errors in the coefficients or an error in the substitution of t = 4. Some tried to use suvat formulae to find the acceleration and a few integrated rather than differentiated.

Common mistakes

• a = dv/dt = 2 − 7/(2√t), so a = 0.25 m s⁻² at t = 4
• integrating instead of differentiating
• suvat cannot be used (acceleration not constant)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), successful candidates integrated the expression for v with respect to t and were usually able to obtain a correct, unsimplified expression for the displacement of the particle at time t. There were then occasional errors with the simplification often due to incorrect coefficients. There was no penalty for not including a constant of integration. Successful candidates then used the limits of t = 1 and t = 2 to calculate the distance XY. A surprising number of candidates thought that finding XY meant that they then had to multiply these two values. Since there was a printed answer, candidates needed to clearly show all stages of their working including the use of surds to get to the required form of the answer. Of those that tried to use surds, some struggled to reach the given answer whilst others did not show enough working to gain the final A mark. Ignoring the instructions at the top of the question, many candidates used calculators and decimals appeared in their solutions which resulted in the solutions being incomplete and losing the final mark. Very few candidates used differentiation in this part.

Common mistakes

• integrate v: t² − (14/3)t^(3/2) + 6t
• XY = s(2) − s(1) (no change of direction so displacement = distance)
• keep surds: 2^(3/2) = 2√2 to reach (41 − 28√2)/3
• decimals lose the final mark
• multiplying the two values

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(b) The majority of candidates realised the need to differentiate the given expression for r, with c = 3 substituted, to obtain the velocity, and most of these differentiated correctly. They then substituted t = 4 and used Pythagoras to find the speed, giving their answer either as an exact value or an appropriate decimal equivalent. A minority correctly found the velocity but then stopped and lost two marks. A few candidates delayed substituting c = 3 until the end and a few delayed putting t = 4 which made the working a lot more demanding. About half worked with column vectors instead of using i - j form. Common errors included calculation slips when evaluating the speed or when substituting t = 4, misreading the expression such as losing the negative sign in the j-component and losing the i and j and not recovering to use components. A few candidates tried to use a suvat approach not realising that this was not a constant acceleration situation. (c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• differentiate r (powers of t decrease by 1): v = (3/2)t^(−1/2)i − (3/4)tj
• put t = 4 in both components then Pythagoras: speed = 3√17/4 ≈ 3.09 m s⁻¹
• stopping at the velocity vector
• suvat invalid (acceleration not constant)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(c) Many of those who differentiated in (b), successfully differentiated their velocity vector to obtain the acceleration vector. However, a significant number of candidates did not know how to proceed from there and the second stage of the process proved to be a discriminator. It was only the stronger candidates who realised that their acceleration vector had to be a multiple of the given vector. Those who did make this link generally were able to complete the question successfully. A number of unsuccessful candidates just set the i-component of the acceleration equal to -1 (or more rarely to -27). Most of k -- those who compared a to ( 27 ) i j were able to reach an equation in t only which they then tried to solve. Fewer candidates used a ratio approach but a significant minority of these used the reciprocal of the correct ratio but were still able to score the method mark. Having reached a correct equation in t or T only, some then struggled to obtain the correct answer. There were a few candidates who tried to use a suvat approach for this part not realising that this was not a constant acceleration situation and scored nothing.

Common mistakes

• a = −(3/4)t^(−3/2)i − (3/4)j must be a multiple of (−i − 27j): use the ratio of components (not the reciprocal)
• setting the i component equal to −1 is wrong
• T = 9

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), virtually all the candidates achieved the mark either by substitution of t = 3 to show that v = 0 or by putting v = 0 and solving for t. A very small number of candidates failed to reject the second negative solution of their quadratic. Most candidates differentiated correctly

Common mistakes

• substitute t = 3 into v = 15 − t² − 2t to get 0 (or solve v = 0 and reject the negative root)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

- but there were a few who rearranged their v = 0 equation and in part (b) and then obtained 8 - not understanding then differentiated and they got 2t+2 as their derivative. Many stopped at 8 that magnitude had to be positive. A few tried to do some sort of “magnitude” calculation using Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c),

Common mistakes

• a = dv/dt = −2t − 2, at t = 3 a = −8
• magnitude 8 must be positive
• Pythagoras-style "magnitude" calculations are wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c), most integrated correctly and scored the first two marks. Very few got the differentiating and integrating the wrong way round this year. But after that, the majority just found the value of the integral at t = 4 (with or without subtracting the value at t = 0). Some realised that they had to split the integral into from 0 to 3 then 3 to 4 but several of these then added the two components instead of subtracting them. Those who drew a v-t graph were generally more successful when seeing the area below the axis seemed to help. There were relatively few correct answers to this part.

Common mistakes

• integrate to 15t − t³/3 − t² (must be seen: calculator-only answers score 0)
• P turns at t = 3, so distance = s(3) + (s(3) − s(4)) = 94/3 m
• most just evaluated at t = 4
• adding the two parts instead of subtracting the negative one
• a v-t sketch helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally a well answered question. Some found the velocity vector at t = 0 and went no further, either forgetting, or not being aware that they needed to use Pythagoras’ Theorem to find the speed. A small number of candidates scored the M mark only, if they had an incorrect velocity vector but used Pythagoras’ Theorem correctly. The majority of candidates who attempted the second part used a correct method, setting the i and j components of the velocity equal to each other or both equal to a constant which was then eliminated. A few also successfully used trial and error. However, a significant number set both i and j coefficients equal to 1, solved one or two separate equations and scored nothing.

Common mistakes

• differentiate each component of v
• answer must be a vector (2t − 3)i + 4tj

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally a well answered question. Some found the velocity vector at t = 0 and went no further, either forgetting, or not being aware that they needed to use Pythagoras’ Theorem to find the speed. A small number of candidates scored the M mark only, if they had an incorrect velocity vector but used Pythagoras’ Theorem correctly. The majority of candidates who attempted the second part used a correct method, setting the i and j components of the velocity equal to each other or both equal to a constant which was then eliminated. A few also successfully used trial and error. However, a significant number set both i and j coefficients equal to 1, solved one or two separate equations and scored nothing.

Common mistakes

• perpendicular to i means the i-component of acceleration is zero: 2t − 3 = 0, t = 1.5
• setting both components to zero scores M0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally well answered by candidates with accurate and efficient methods frequently shown. The majority knew to differentiate to obtain an expression for velocity and were able to do so correctly and differentiating again to find the acceleration was almost always done correctly also. Some didn't set their velocity equal to zero and instead put their acceleration equal to zero, commonly scoring 3 out of the 6 marks available in this part. A number of candidates integrated instead of differentiating but this was rare. A tiny minority attempted to use suvat formulae in this variable acceleration question. Part (b) was much more challenging and most candidates didn't see the connection with the work they'd done in part (a) and what this meant for the direction of travel and total distance travelled. The majority of the successful candidates showed three separate sets of working for each of the three sections and then knew they needed to add the modulus of each section. This was often supported by sketching a graph. Integrating the velocity between 0 and 4 or using the given equation and a single calculation between 0 and 4 was the most common error leading to an answer of 16/3. A significant number of candidates did find s2, s3 and s4 (and often s1 as well) but then just added these values together so just scored the first method mark in this part.

Common mistakes

• v = ds/dt = t² − 5t + 6
• at rest means v = 0 (not a = 0), t = 2 or 3
• a = 2t − 5 gives 1 and −1
• integrating or suvat score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

attempted to use suvat formulae in this variable acceleration question. Part (b) was much more challenging and most candidates didn't see the connection with the work they'd done in part (a)

Common mistakes

• P changes direction at t = 2 and t = 3 from (a)
• distance = s₂ + (s₂ − s₃) + (s₄ − s₃) = 17/3 (5.7), not the displacement s₄ = 16/3
• most did not see the link to (a)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

and went on to use Pythagoras correctly to find an exact answer or an answer correct to the i and j and then finding 12-6 sqrt 2 rather than apply Pythagoras. A small number 2sf or better. Common mistakes were to stop once the velocity had been found, losing subtracted the squared values rather than adding in their Pythagoras. In the second part, most candidates identified that this was variable acceleration and differentiated correctly to find the acceleration vector. Common mistakes were losing either or both i and j and not recovering, integrating instead of differentiating and, despite the question specifying the answer was to be given in i j form, final answers given in column vector form were occasionally seen. Many candidates integrated to achieve the first two marks in part (c),

Common mistakes

• variable acceleration so differentiate v (not integrate, not suvat)
• keep both i and j
• answer 6ti − 3t^(−1/2)j in i, j form, not a column vector

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

occasionally seen. Many candidates integrated to achieve the first two marks in part (c), but not all used the displacement at t = 4 to find the constant of integration. Those that found a vector equation containing a constant of integration C, went on to find C and then substituted t = 1 to reach the correct answer although some responses contained numerical errors involving the negative signs. A common error was to use suvat formulae or r = r0 + vt. rather than integrate.

Common mistakes

• integrate v and use r = i − 4j at t = 4 to find the constant vector C
• suvat or r = r₀ + vt is invalid (acceleration not constant)
• sign slips
• answer −62i + 24j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(a) Most candidates knew that they needed to differentiate but integration was also seen once. Some went on to put a = 0, solving to get t = 5 which they went on to use in part (b) but this subsequent working was ignored in part (a).

Common mistakes

• differentiate v to get a = 10 − 2t
• integrating is wrong
• setting a = 0 is not required

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Some went on to put a = 0, solving to get t = 5 which they went on to use in part (b) but this subsequent working was ignored in part (a). (b) The majority used v = 0 when t = 6, found a value of k and obtained a quadratic. Unfortunately, many made sign errors and their expression had +24 instead of -24. If they went on to solve their incorrect quadratic using a correct method, these candidates scored two of the four marks available. There were several completely correct answers.

Common mistakes

• instantaneously at rest means v = 0
• v = 0 at t = 6 gives k = 24
• sign errors (+24 instead of −24)
• solve the quadratic for t = 4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(c) Some candidates knew that they needed to integrate v but a few tried to use suvat equations, losing all the marks and some did not substitute a numerical value for k. The last two marks were rarely scored as the change of direction was not appreciated.

Common mistakes

• integrate v (suvat is invalid: acceleration not constant)
• v changes sign at t = 4 so split the integral 0 to 4 and 4 to 6 and add magnitudes: 116/3 m
• substitute k = 24 first

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was well answered by most candidates who realised that they needed to differentiate although a few candidates did omit the vector notation and therefore achieved zero marks. Part (b) proved to be the least popular part of this question.

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• keep the vector notation (M0 if i, j disappear)
• answer (3/2)t^(−½) i − 2j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

achieved zero marks. Part (b) proved to be the least popular part of this question. 1 = 2t and then 3t Successful candidates employed a ratio method to obtain the equation 2 solved it. However, a significant minority of candidates incorrectly equated the i and j parts and then solved the two equations to obtain the correct answer. Candidates should note that this approach scored no marks. Other errors included using the acceleration vector or the displacement vector rather than the velocity vector when dealing with direction of travel. Part (c) tended to be well answered with a good number of candidates

Common mistakes

• moving in direction i − j means the i and j components of v are equal and opposite: 3t^½ = 2t
• equating v to i − j scores nothing
• use v, not a or r
• t = 9/4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

direction of travel. Part (c) tended to be well answered with a good number of candidates integrating and gaining 2 or 3 marks here. It was disappointing to see a few using constant acceleration formulae. Part (d) allowed those who had made little progress on the first 3

Common mistakes

• r = ∫v dt, both powers increase by 1
• use r = −j at t = 1 to get C = −2i
• suvat is invalid (acceleration not constant)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

acceleration formulae. Part (d) allowed those who had made little progress on the first 3 parts, to score some marks as it only required the use of the given velocity vector to get started. For those who attempted it, the use of Pythagoras’ theorem on the velocity components was very accessible and a number of candidates solved the quadratic to obtain t = 4. Sign errors here were common, however, with many ending up with a wrong t value. Some candidates then scored a further two M marks for substituting their value of t into their position vector and then applying Pythagoras to find the distance. The question required an exact distance and so only surd answers earned the final mark.

Common mistakes

• speed 10 means 9t + 4t² = 100, t = 4 (sign errors common)
• substitute into r: 14i − 16j
• distance is the magnitude of r
• exact answer 2√113 (√452), decimals lose the final mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), a significant minority misunderstood the question and solved v = 0 instead of a = 0. The majority, however, did multiply out and differentiate correctly to find a before then equating it to zero. A few errors in multiplying out the brackets were seen and a significant number had changed the sign of v which then had consequences in part (b).The majority, for part (b), realised that the particle the answer. A few, however, used the t value that had been found in part(a) and lost the method mark.

Common mistakes

• multiply out v = 14 + 3t − 2t² and differentiate
• stops accelerating means a = 0 (not v = 0), t = 3/4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

had consequences in part (b).The majority, for part (b), realised that the particle would change its direction of motion when v = 0 and solved this equation to give t = 3.5. They were then able to integrate v correctly and use their t value to obtain the answer. A few, however, used the t value that had been found in part(a) and

Common mistakes

• changes direction when v = 0, t = 3.5 (not the t from (a))
• integrate v (suvat scores M0)
• s = 931/24 = 38.8 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• variable acceleration so suvat scores nothing
• integrate each component and include the constant 36i
• v = 8i − (28/3)j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• perpendicular to i means the i component of v is zero: t − 2t² + 36 = 0 gives t = 4.5
• equating the j component to zero (or i to j) scores nothing
• use velocity, not displacement

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (i) involved variable acceleration but a significant minority of candidates tried to apply suvat equations and so achieved no credit. In part (i)(a), many did complete the required integration to find an expression for v and substitute t = 4; a fair number of correct answers were seen with the main errors either being numerical slips or neglecting to include the constant, 36i. In part (i)(b), candidates needed to realise that moving perpendicular to i implies the i component of v is zero; those who equated the j components to zero achieved no marks. Occasionally the i and j components were equated and some candidates did not know how to proceed at all. Some focussed on the displacement vector. Those who knew what to do generally obtained the correct answer. Part (ii) was generally found more accessible and many of the candidates who had used suvat in (i) now realised that they had to differentiate here to find the velocity. Most knew how to use the given speed to create an equation in t, although a few just tried to equate their vector to 5 and then either gave up or just dropped the i and j. A fair number achieved full marks for this part of the question.

Common mistakes

• differentiate r to get v = (2t − 1)i + 3j
• speed is the magnitude: (2t − 1)² + 3² = 5² gives t = 2.5
• equating the vector to 5 or dropping i and j is wrong
• not √(x² − y²)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In the first part, the majority of candidates knew that finding acceleration involved differentiating the given expression for the velocity and, apart from the occasional slip, this was generally done well. However, a fair number then found either t when a = 0 or a when t = 0 without realising that, to find the relevant value of t, it was necessary to solve the quadratic equation obtained by setting v = 0 (when the particle is at instantaneous rest) . Many of those with the correct working lost the final mark for giving a negative answer for the ‘magnitude’ of the acceleration. Most candidates knew they had to integrate for part (b) which they did mostly successfully and many

Common mistakes

• instantaneous rest means v = 0: solve 12 + 4t − t² = 0 to t = 6 (reject −2)
• a = dv/dt = 4 − 2t = −8
• magnitude 8 m s⁻² (positive)
• not a when t = 0 or t when a = 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most candidates knew they had to integrate for part (b) which they did mostly successfully and many went on to substitute t = 3 as required to find the distance travelled. There were a few numerical slips in carrying out the substitution but there were many correct solutions seen to this part of the question.

Common mistakes

• integrate v: s = 12t + 2t² − t³/3
• v > 0 for 0 ≤ t ≤ 3 so distance = s(3) = 45 m

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In the first part, the majority of candidates attempted to differentiate the given vector expression for velocity to find the acceleration as required. This was completed mostly successfully with just the occasional slip. Occasionally the expression was integrated or the fact that it was a vector just ignored with the coefficients of i and j being added; however, such instances were relatively rare. Those who completed the differentiation generally substituted t = 4 to obtain the acceleration and many correct answers were seen. Some went on to calculate the magnitude which was not actually required here but generally the marks had already been achieved for a correct vector. Some candidates worked in column vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration in part (a).

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• do not integrate or add the i and j coefficients
• answer 6i − 15j as a vector (finding the magnitude is not needed)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration was required in order to find the position vector. Mostly the indices were dealt with correctly although errors in the coefficients were more common. Some stopped there without either using the initial conditions or substituting t = 4. Many failed to achieve the final mark as a result of numerical/sign errors. A minority attempted to use suvat formulae to solve the problem despite having used calculus

Common mistakes

• r = ∫v dt then add r₀ = −20i + 20j
• coefficient slips (3t² and 2t^(5/2))
• suvat is invalid as acceleration is not constant
• answer 28i − 44j

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), many students realised they had to find the velocity to determine when the particle is at instantaneous rest and attempted to differentiate the given expression for x. Some failed to multiply out the brackets first and made no valid progress. Nevertheless, a fair number derived the relevant cubic and solved it successfully to find the 3 values for t. On occasions, t was cancelled out from the equation and the solution t = 0 was lost.

Common mistakes

• multiply out first: x = ½(t⁴ − 2t³ + t²), v = 2t³ − 3t² + t
• v = 0 gives t(2t − 1)(t − 1) = 0 so t = 0, ½, 1
• cancelling t loses t = 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), a lack of understanding of the difference between distance and displacement became very apparent with the vast majority of students substituting only t = 2 (and sometimes t = 0) in an attempt to find the distance travelled in the first 2 seconds. Surprisingly, many integrated their expression for v rather than using the one for x that was given in the question (or even integrated the expression for x). Very few used their answers to (a) to indicate changes in direction and thereby calculate the whole distance. Part (c) also proved to be a challenge with few students scoring either of the two available

Common mistakes

• distance, not displacement: x = 0, 1/32, 0, 2 at t = 0, ½, 1, 2, so distance = 1/32 + 1/32 + 2 = 2 1/16 m
• substituting only t = 2 is the common error
• use the given x rather than integrating v

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The vast majority of students recognised the requirement to integrate to find the position vector and subsequently to substitute the given values of t. Occasionally, these were substituted straight into the velocity vector but such instances were rare and much correct working was seen. Most proceeded to find the magnitude of the vector joining the two points to give the exact distance as a surd as required; only a very small minority went straight to a rounded decimal value.

Common mistakes

• integrate v to get r = 2t^½ i − 2t² j (+ C)
• positions at t = 1 and t = 4, not substituting into v
• exact distance √904 = 2√226, not a rounded decimal