A2 June 2022 Q2
2. A cyclist and her cycle have a combined mass of 60 kg. The cyclist is moving along a straight horizontal road and is working at a constant rate of 200 W.
When she has travelled a distance \(x\) metres, her speed is \(v\ \text{m s}^{-1}\) and the magnitude of the resistance to motion is \(3v^2\) N.
The distance travelled by the cyclist as her speed increases from \(2\ \text{m s}^{-1}\) to \(4\ \text{m s}^{-1}\) is \(D\) metres.
| Scheme | Marks | AO |
|---|---|---|
| Use of \(P = Fv\) | B1 | 3.3 |
| Equation of motion \(\left(F - 3v^2 = 60a\right)\) | M1 | 2.1 |
| \(\dfrac{200}{v} - 3v^2 = 60v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | A1 | 2.5 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{200 - 3v^3}{60v^2}\) * | A1* | 2.2a |
| (4) |
Notes
B1: Seen or implied
Not just quoted. Need at least \(200 = Fv\)
Could be on its own, in an equation or on a diagram
M1: Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors
Allow with \(m\) not substituted
A1: Correct equation - any equivalent form with correct acceleration
A1*: Obtain given answer from correct working
Must be as written in the question but could swap LHS and RHS
| Scheme | Marks | AO |
|---|---|---|
| \(\Rightarrow \displaystyle\int \dfrac{60v^2}{200 - 3v^3}\,\mathrm{d}v = \int 1\,\mathrm{d}x \qquad \left(-\dfrac{60}{9}\ln\left(200 - 3v^3\right) = x(+C)\right)\) | M1 | 1.1b |
| \(D = \left[-\dfrac{60}{9}\ln\left(200 - 3v^3\right)\right]_2^4 = -\dfrac{60}{9}\ln\left(\dfrac{200 - 3\times 64}{200 - 3\times 8}\right)\) | M1 | 1.1b |
| \(= \dfrac{60}{9}\ln\dfrac{176}{8} = \dfrac{60}{9}\ln 22\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Separate variables and integrate to obtain \((x =)\,k\ln(\ldots\ldots)\)
(Constant of integration not required)
Condone if the \(x\) is not explicitly stated but M0 if it is an incorrect function.
M1: Use limits correctly in an expression containing \(k\ln\left(200 - 3v^3\right)\) to find \(D\)
Substitute and subtract in the correct order
A1: Obtain exact answer from correct working
Any equivalent single term
No working seen is Max M1M0A0