A-Level Mathematics October/November 2024 Q7(b): The car now travels at a constant speed up a hill inclined at an angle of sin¯¹0.15 to…

1 answer

• ▲0
  accepted ✓

  Let the speed of the car be $v$ m/s. The car is travelling at a constant speed, so the forces acting parallel to the slope are in equilibrium.

  The angle of inclination is $\alpha$, where $\sin \alpha = 0.15$.

  Let the driving force be $D$ N and the resistance to motion be $R$ N. From the context of the question, we assume the mass of the car is $m = 1200$ kg and the resistance is given by $R = 40v$. We take $g=10$ m/s².

  Resolving forces parallel to the slope: $D = R + mg \sin \alpha$

  Substituting the known values: $D = 40v + (1200)(10)(0.15)$ $D = 40v + 1800$

  The greatest possible speed is achieved when the engine is working at its maximum power, $P_{max} = 92160$ W.

  The relationship between power, driving force, and speed is $P = D \times v$.

  $92160 = (40v + 1800)v$ $92160 = 40v^2 + 1800v$

  Rearranging into a quadratic equation: $40v^2 + 1800v - 92160 = 0$

  Divide by 40 to simplify: $v^2 + 45v - 2304 = 0$

  We solve this quadratic equation for $v$ using the quadratic formula, $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $v = \frac{-45 \pm \sqrt{45^2 - 4(1)(-2304)}}{2(1)}$ $v = \frac{-45 \pm \sqrt{2025 + 9216}}{2}$ $v = \frac{-45 \pm \sqrt{11241}}{2}$

  Since speed $v$ must be positive, we take the positive root: $v = \frac{-45 + \sqrt{11241}}{2}$ $v = \frac{-45 + 106.02358...}{2}$ $v = \frac{61.02358...}{2}$ $v = 30.51179...$

  Therefore, the greatest possible speed is $30.5$ m/s (3 s.f.).

  ---

  How the marks are awarded

  • M1 — For resolving forces parallel to the slope to find a two-term expression for the driving force. The model answer correctly identifies the resistance force ($40v$) and the component of weight ($1200 \times 10 \times 0.15$) acting against the driving force, giving $D = 40v + 1800$.
  • DM1 — For using the maximum power $P = 92160$ and the relationship $P = D \times v$ to form an equation in terms of the speed, $v$, only. The model answer correctly forms the equation $92160 = (40v + 1800)v$.
  • DM1 — For correctly rearranging the equation into a three-term quadratic and making a valid attempt to solve it for $v$. The model answer shows the simplified quadratic $v^2 + 45v - 2304 = 0$ and correctly applies the quadratic formula.
  • A1 — For obtaining the correct final answer $30.51179...$. The model answer calculates this value and rounds correctly to 3 significant figures.

  Common mistakes

  • Forgetting one of the forces acting down the slope, for example, only including the resistance or only the weight component, leading to an incorrect expression for the driving force.
  • Using cosine instead of sine for the component of weight parallel to the slope (i.e., calculating $mg \cos \alpha$).
  • Incorrectly applying the power formula, for example by setting Power = Driving Force, or by calculating the driving force for a fixed speed instead of leaving speed as a variable.
  • Making an error when solving the quadratic equation, such as a sign error in the formula, or choosing the non-physical negative root for the speed.

  Examiner tip: Master the relationship between forces, motion, and power by first applying Newton's Laws to find the driving force, then connecting it to speed and power with the equation $P = Fv$.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.