M2 June 2016 Q1

EdexcelOld spec13 marksVariable Acceleration (Calculus)

1. A particle \(P\) moves along a straight line. The speed of \(P\) at time \(t\) seconds \((t \geqslant 0)\) is \(v\) m s\(^{-1}\), where \(v = (pt^2 + qt + r)\) and \(p\), \(q\) and \(r\) are constants. When \(t = 2\) the speed of \(P\) has its minimum value. When \(t = 0\), \(v = 11\) and when \(t = 2\), \(v = 3\)

Find

(a) the acceleration of \(P\) when \(t = 3\) (8)

(b) the distance travelled by \(P\) in the third second of the motion. (5)

Mark scheme (a)

Scheme	Marks
\(t = 0,\ v = 11\ \Rightarrow r = 11\)	B1
\(t = 2,\ v = 3\ \Rightarrow 4p + 2q + 11 = 3,\)	M1
\(4p + 2q = -8\)	A1
Differentiate to find acceleration	M1
\(a = 2pt + q\)	A1
\(t = 2,\ a = 0\ \Rightarrow 4p + q = 0\)	DM1
\(\Rightarrow -q + 2q = -8,\ \ q = -8,\ \ p = 2\) \(\left(v = 2t^2 - 8t + 11\right)\)	A1
\(t = 3,\ \ a = 4t - 8 = 4\) (ms\(^{-2}\))	A1
	(8)

Notes

M1 Accept \(4p + 2q + r = 3\)

A1 Any equivalent unsimplified form with 11 used

M1 OR use symmetry, \(t = 4, v = 11\)

A1 \(\Rightarrow 11 = 16p + 4q + 11,\ \ 4p + q = 0\)

DM1 2^nd eqn in \(p\) & \(q\) and solve for \(p\) & \(q\). Dependent on both previous m marks

1a alt

Min speed at \(t = 2\ \Rightarrow\) \(v = \left(pt^2 + qt + r\right) = k(t - 2)^2 + c\)	B1 M1
\(v = k(t - 2)^2 + 3\)	A1
\(t = 0,\ v = 11\ \Rightarrow 4k + 3 = 11,\)	M1
\(k = 2\)	A1
Differentiate to find acceleration	DM1
\(a = 4(t - 2)\)	A1
\(t = 3,\ \ a = 4\) (m s\(^{-2}\))	A1

M1 Completed square form.

A1 Correct completed square form

M1 Solve for \(k\)

A1 \(v = 2(t - 2)^2 + 3\left(= 2t^2 - 8t + 11\right)\)

DM1 Dependent on both previous m marks

Mark scheme (b)

Scheme	Marks
Integrate: \(\displaystyle\int 2(t - 2)^2 + 3\,\mathrm{d}t = \dfrac{2}{3}(t - 2)^3 + 3t\ (+C)\) or \(\displaystyle\int 2t^2 - 8t + 11\,\mathrm{d}t = \dfrac{2}{3}t^3 - 4t^2 + 11t\ (+C)\)	M1
At most one error seen	A1ft
All correct	A1ft
\(\left[\dfrac{2}{3}(t - 2)^3 + 3t\right]_2^3 = \left(\dfrac{2}{3} + 9\right) - (0 + 6)\) or \(\left[\dfrac{2}{3}t^3 - 4t^2 + 11t\right]_2^3\) \(= (18 - 36 + 33) - \left(\dfrac{16}{3} - 16 + 22\right)\)	DM1
\(3\dfrac{2}{3}\) (m)	A1
	(5)
	(13 marks)

Notes

M1 follow their coefficients found in (a). Accept in \(p\), \(q\), \(r\)

A1ft For their coefficients

A1ft For their coefficients provided \(\neq 0\)

DM1 Use of \(t = 2, t = 3\) as limits on a definite integral (or subtract distances to cancel \(C\)). Dependent on having integrated. Allow with \(p\), \(q\), \(r\)

A1 Accept exact equivalent or 3.7 or better