A2 June 2025 Q5

EdexcelCurrent spec10 marksVariable Force & Kinematics

5. A car of mass 1000 kg is moving along a straight horizontal road.
The engine of the car produces a constant driving force of 800 N.
At time \(t\) seconds, \(t \geqslant 0\), the speed of the car is \(v\ \text{m s}^{-1}\) and the resistance to the motion of the car has magnitude \(2v^2\) N.

The time taken for the speed of the car to increase from \(5\ \text{m s}^{-1}\) to \(15\ \text{m s}^{-1}\) is \(T\) seconds.

(a) Show that \(T = \dfrac{25}{2}\ln\left(\dfrac{21}{5}\right)\) (7)

When \(t = 0\), \(v = 2\)

(b) Show that \(\dfrac{20 + v}{20 - v} = p\mathrm{e}^{qt}\) where \(p\) and \(q\) are rational numbers to be found. (2)

Mark scheme (a)

Scheme	Marks	AO
Equation of motion	M1	3.4
\(800 - 2v^2 = 1000\dfrac{\mathrm{d}v}{\mathrm{d}t}\)	A1 A1	1.1b 1.1b
\(\Rightarrow \displaystyle\int \dfrac{500}{400 - v^2}\,\mathrm{d}v = \int 1\,\mathrm{d}t\)	M1	2.1
\(\dfrac{25}{2}\ln\left\|\dfrac{20 + v}{20 - v}\right\| = t\ (+C)\)	A1	1.1b
\(T = \left[\dfrac{25}{2}\ln\left(\dfrac{20 + v}{20 - v}\right)\right]_5^{15}\)	M1	1.1b
\(= \dfrac{25}{2}\ln\left(\dfrac{35}{5} \times \dfrac{15}{25}\right) = \dfrac{25}{2}\ln\left(\dfrac{21}{5}\right)\) *	A1*	2.2a
	(7)

Notes

M1: Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors

A1: Unsimplified equation in \(v\) and \(t\) with at most one error

A1: Correct unsimplified equation in \(v\) and \(t\)

M1: Separate variables and integrate to a correct logarithmic form.
May see attempts using substitution / trig etc. but need to get to logarithmic form.

A1: Any equivalent form. Constant of integration and modulus signs not required.

M1: Use correct limits correctly in an expression containing \(\lambda\ln(20 - v)\) and \(\mu\ln(20 + v)\)

A1*: Obtain given answer from correct working (condone disappearance of “\(\mathrm{d}v\)” or “\(\mathrm{d}t\)” on integrals)

Mark scheme (b)

Scheme	Marks	AO
\(t = 0,\ v = 2\ \Rightarrow\ C = \dfrac{25}{2}\ln\dfrac{22}{18}\ \left(= \dfrac{25}{2}\ln\dfrac{11}{9}\right)\)	M1	3.1b
\(\left(\dfrac{20 + v}{20 - v}\right) = \dfrac{11}{9}\mathrm{e}^{\frac{2t}{25}}\) *	A1*	2.2a
	(2)

Notes

M1: Use boundary condition to find constant of integration for their integral.

A1*: Obtain given conclusion from correct working. Allow any exact equivalent for \(p\) & \(q\).

Mark scheme (c)

Scheme	Marks	AO
As \(\mathrm{e}^{qt} \gt 0\), \(v \lt 20\) *	B1ft *	2.4
	(1)
	(10 marks)

Notes

B1*: Obtain given conclusion. Follow their \(p \gt 0\) but must be referring to the idea that the expression must be positive because it is an exponential function.