October 2020 Paper 3 Mechanics Q3

EdexcelCurrent spec12 marksVariable Acceleration (Calculus)Vectors

(i) At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves so that its acceleration \(\mathbf{a}\ \text{m s}^{-2}\) is given by\[\mathbf{a} = (1 - 4t)\,\mathbf{i} + (3 - t^2)\,\mathbf{j}\]At the instant when \(t = 0\), the velocity of \(P\) is \(36\mathbf{i}\ \text{m s}^{-1}\)

(a) Find the velocity of \(P\) when \(t = 4\) (3)

(b) Find the value of \(t\) at the instant when \(P\) is moving in a direction perpendicular to \(\mathbf{i}\) (3)

(ii) At time \(t\) seconds, where \(t \geqslant 0\), a particle \(Q\) moves so that its position vector \(\mathbf{r}\) metres, relative to a fixed origin \(O\), is given by\[\mathbf{r} = (t^2 - t)\,\mathbf{i} + 3t\,\mathbf{j}\]Find the value of \(t\) at the instant when the speed of \(Q\) is \(5\ \text{m s}^{-1}\) (6)

Mark scheme (i)(a)

Scheme	Marks	AO
Integrate \(\mathbf{a}\) wrt \(t\) to obtain velocity	M1	3.4
\(\mathbf{v} = (t - 2t^2)\mathbf{i} + \left(3t - \dfrac{1}{3}t^3\right)\mathbf{j}\ (+\mathbf{C})\)	A1	1.1b
\(8\mathbf{i} - \dfrac{28}{3}\mathbf{j}\ (\text{m s}^{-1})\)	A1	1.1b
	(3)

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.

Mark scheme (i)(b)

Scheme	Marks	AO
Equate \(\mathbf{i}\) component of \(\mathbf{v}\) to zero	M1	3.1a
\(t - 2t^2 + 36 = 0\)	A1ft	1.1b
\(t = 4.5\) (ignore an incorrect second solution)	A1	1.1b
	(3)

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

Mark scheme (ii)

Scheme	Marks	AO
Differentiate \(\mathbf{r}\) wrt to \(t\) to obtain velocity	M1	3.4
\(\mathbf{v} = (2t - 1)\mathbf{i} + 3\mathbf{j}\)	A1	1.1b
Use magnitude to give an equation in \(t\) only	M1	2.1
\((2t - 1)^2 + 3^2 = 5^2\)	A1	1.1b
Solve problem by solving this equation for \(t\)	M1	3.1a
\(t = 2.5\)	A1	1.1b
	(6)
	(12 marks)

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