M4 June 2016 Q4

4. A particle \(P\) of mass 9 kg moves along the horizontal positive \(x\)-axis under the action of a force directed towards the origin. At time \(t\) seconds, the displacement of \(P\) from \(O\) is \(x\) metres, \(P\) is moving with speed \(v\) m s\(^{-1}\) and the force has magnitude \(16x\) newtons. The particle \(P\) is also subject to a resistive force of magnitude \(24v\) newtons.

(a) Show that the equation of motion of \(P\) is \[9\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 24\dfrac{\mathrm{d}x}{\mathrm{d}t} + 16x = 0\] (4)

It is given that the general solution of this differential equation is \[x = \mathrm{e}^{-\frac{4}{3}t}\left(At + B\right)\] where \(A\) and \(B\) are arbitrary constants.

When \(t = \dfrac{3}{4}\), \(P\) is travelling towards \(O\) with its maximum speed of \(8\mathrm{e}^{-1}\) m s\(^{-1}\) and \(x = d\).

(b) Find the value of \(d\). (3)

Mark scheme (a)

Scheme	Marks
NL2: \(9\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -24v - 16x\)	M1
	A1
\(9\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 24\dfrac{\mathrm{d}x}{\mathrm{d}t} + 16x = 0\)	M1
	A1
	(4)

Notes

M1 Requires all 3 terms but condone sign errors. Condone \(\dot{x}\) for \(v\). Must be dimensionally correct.

A1 Correct unsimplified equation with \(v\). Accept with \(9a\) in which case accept \(\pm\)

M1 Substitute for \(v\) (seen anywhere) to form equation in \(x\) and \(t\) only

A1 Given answer as printed - from correct solution.

NB: If never see \(v\) used, max score 1/4

Mark scheme (b)

Scheme	Marks
\(\ddot{x} = 0 \Rightarrow 16x = -24\dot{x}\)	M1
\(16d = 24\times 8\mathrm{e}^{-1}\)	A1
\(d = 12\mathrm{e}^{-1}\)	A1
	(3)

Notes

M1 \(\ddot{x} = 0\) used. Accept equivalent forms

A1 \(\dot{x}\) substituted correctly

A1 4.4 or better

(b) alt

	M1
\(x = \mathrm{e}^{-\frac{4}{3}t}(8t + 6)\)	A1
\(d = 12\mathrm{e}^{-1}\)	A1
	(3)

M1 Differentiate twice and find \(A\) and \(B\). Condone use of \(t = \dfrac{3}{4},\ \dot{x} = 8\mathrm{e}^{-1}\)

Mark scheme (c)

Scheme	Marks
\(\dot{x} = -\dfrac{4}{3}\mathrm{e}^{-\frac{4}{3}t}(At + B) + A\mathrm{e}^{-\frac{4}{3}t}\)	M1
	A1
\(-8\mathrm{e}^{-1} = -\dfrac{4}{3}\mathrm{e}^{-1}\left(\dfrac{3}{4}A + B\right) + A\mathrm{e}^{-1}\)	M1
\(-8 = -A - \dfrac{4}{3}B + A\)
\(B = 6\)	A1
The first 4 marks for (c) are available when seen
\(\left(x = \mathrm{e}^{-\frac{4}{3}t}(At + 6)\right)\)
\(t = 0\quad x = 6\)	B1ft
	(5)
	(12 marks)

Notes

M1 Differentiate the given general solution using the product rule

A1 Correct unsimplified

M1 Use \(t = \dfrac{3}{4},\ \dot{x} = -8\mathrm{e}^{-1}\)

B1ft their \(B\)