M2 June 2018 Q6

EdexcelOld spec14 marksVariable Acceleration (Calculus)Vectors

6. A particle \(P\) of mass 0.5 kg moves under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds, \(t \geqslant 0\), \(P\) has velocity \(\mathbf{v}\) m s\(^{-1}\), where\[\mathbf{v} = (4t - 3t^2)\mathbf{i} + (t^2 - 8t - 40)\mathbf{j}\]

(a) Find

(i) the magnitude of \(\mathbf{F}\) when \(t = 3\)

(ii) the acceleration of \(P\) at the instant when it is moving in the direction of the vector \(-\mathbf{i} - \mathbf{j}\). (9)

When \(t = 1\), \(P\) is at the point \(A\). When \(t = 2\), \(P\) is at the point \(B\).

(b) Find, in terms of \(\mathbf{i}\) and \(\mathbf{j}\), the vector \(\overrightarrow{AB}\). (5)

Mark scheme (a)

Scheme	Marks
Differentiate \(\mathbf{v}\): \(\ \ \mathbf{a} = (4 - 6t)\mathbf{i} + (-8 + 2t)\mathbf{j}\)	M1A1
Use of \(\mathbf{F} = m\mathbf{a}\) and substitute \(t = 3\): \(\mathbf{F} = 0.5\left((4 - 6 \times 3)\mathbf{i} + (-8 + 2 \times 3)\mathbf{j}\right) = -7\mathbf{i} - \mathbf{j}\)	DM1
Use of Pythagoras’ theorem:	DM1
\(\|\mathbf{F}\| = \sqrt{49 + 1} = \sqrt{50}\left(= 5\sqrt{2} = 7.07...\right)\)	A1
For \(\mathbf{v}\), \(\mathbf{i}\) component= \(\mathbf{j}\) component: \(\left(4t - 3t^2\right) = \left(-40 - 8t + t^2\right)\)	M1
Solve for \(t\): \(4t^2 - 12t - 40 = 0,\ \Rightarrow t^2 - 3t - 10 = 0\) \((t - 5)(t + 2) = 0,\ \ t = 5\)	DM1 A1
\(\mathbf{a} = (4 - 30)\mathbf{i} + (-8 + 10)\mathbf{j} = -26\mathbf{i} + 2\mathbf{j}\) (ms\(^{-2}\))	A1
	(9)

Notes

M1A1 Anywhere in (a)

DM1 Dependent on the first M1

NB Could use Pythagoras and then use \(\mathbf{F} = m\mathbf{a}\). 1^st M1 – 1^st step. 2^nd M1 - 2^nd step

A1 7.1 or better

M1 With no incorrect equations in \(t\) seen

DM1 Dependent on the previous M, Must see method if solving an incorrect quadratic. A1 Only - could be implied by later rejection of -2

A1 Only

Mark scheme (b)

Scheme	Marks
Integrate \(\mathbf{v}\): \(\mathbf{r} = \left(2t^2 - t^3(+p)\right)\mathbf{i} + \left(-40t - 4t^2 + \dfrac{1}{3}t^3(+q)\right)\mathbf{j}\)	M1 A2
\(\mathbf{r}_1 = \mathbf{i} - 43\dfrac{2}{3}\mathbf{j},\ \ \mathbf{r}_2 = -93\dfrac{1}{3}\mathbf{j}\ \ \ \ \ \ \overrightarrow{AB} = \mathbf{r}_2 - \mathbf{r}_1\)	DM1
\(\overrightarrow{AB} = -\mathbf{i} - 49\dfrac{2}{3}\mathbf{j}\left(= -\mathbf{i} - \dfrac{149}{3}\mathbf{j}\right)\)	A1
	(5)
	(14 marks)

Notes

A2 -1 ee

DM1 \(\left(\dfrac{131}{3},\ \dfrac{280}{3}\right)\) Use limits in a definite integral or to evaluate a constant of integration. Dependent on the previous M1

A1 49.7 or better