AS June 2024 Q3

EdexcelCurrent spec11 marksVariable Force & Kinematics

3. A particle \(P\) is moving along the \(x\)-axis. At time \(t\) seconds, \(P\) has velocity \(v\ \text{m s}^{-1}\) in the positive \(x\) direction and acceleration \(a\ \text{m s}^{-2}\) in the positive \(x\) direction.

In a model of the motion of \(P\)

\[a = 4 - 3v\]

When \(t = 0\), \(v = 0\)

(a) Use integration to show that \(v = k\left(1 - \mathrm{e}^{-3t}\right)\), where \(k\) is a constant to be found. (7)

When \(t = 0\), \(P\) is at the origin \(O\)

(b) Find, in terms of \(t\) only, the distance of \(P\) from \(O\) at time \(t\) seconds. (4)

Mark scheme (a)

Scheme	Marks	AO
\(a = 4 - 3v \quad \Rightarrow \quad \displaystyle\int \dfrac{1}{4-3v}\,\mathrm{d}v = \int 1\,\mathrm{d}t\)	M1	2.1
Integrate both sides of the equation	M1	1.1b
\(\Rightarrow -\dfrac{1}{3}\ln\|4-3v\| = t\ (+C)\)	A1	1.1b
Use \(t = 0\), \(v = 0\)	M1	3.4
\(t = \dfrac{1}{3}\ln\left(\dfrac{4}{4-3v}\right)\)	A1	1.1b
\(\Rightarrow \mathrm{e}^{3t} = \dfrac{4}{4-3v},\quad 4 - 3v = 4\mathrm{e}^{-3t}\)	M1	1.1b
\(v = \dfrac{4}{3}\left(1 - \mathrm{e}^{-3t}\right)\) *	A1*	2.2a
	(7)

Notes

M1: Use \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t}\) and separate the variables to form integrals in \(v\) and \(t\)

M1: Integrate to obtain terms \(p\ln(4-3v)\) and \(qt\)

A1: Or equivalent. Accept with brackets in place of modulus signs. Condone missing constant of integration

M1: Use boundary conditions in the model to evaluate constant of integration or as limits on a definite integral

A1: Or equivalent

M1: Rearrange to obtain expression for \(v\) in terms of \(t\)

A1*: Obtain given form with \(k = \dfrac{4}{3}\) from correct working

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = k\left(1 - \mathrm{e}^{-3t}\right) \quad \Rightarrow \quad \displaystyle\int 1\,\mathrm{d}x = \int k\left(1 - \mathrm{e}^{-3t}\right)\mathrm{d}t\)	M1	3.3
\(x = k\left(t + \dfrac{1}{3}\mathrm{e}^{-3t}\right)(+C) = \dfrac{4}{3}\left(t + \dfrac{1}{3}\mathrm{e}^{-3t}\right)(+C)\)	M1	1.1b
Use \(t = 0\), \(x = 0\)	M1	3.4
\(x = k\left(t + \dfrac{1}{3}\mathrm{e}^{-3t} - \dfrac{1}{3}\right) = \dfrac{4}{3}\left(t + \dfrac{1}{3}\mathrm{e}^{-3t} - \dfrac{1}{3}\right)\)	A1ft	1.1b
	(4)
	(11 marks)

Notes

M1: Use \(v = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) to form integrals in \(x\) and \(t\)

M1: Integrate to obtain \(\lambda t + \mu\mathrm{e}^{-3t}\ (+C)\)

M1: Use boundary conditions in the model to evaluate constant of integration or as limits on a definite integral

A1ft: Any equivalent form. Follow their \(k\)