Problems

Filters
Clear Filters

1 problem found

2016 Paper 3 Q11
D: 1700.0 B: 1484.0
variable force power resistance Newton's second law integration logarithm partial fractions acceleration car motion inequality comparison

A car of mass \(m\) travels along a straight horizontal road with its engine working at a constant rate \(P\). The resistance to its motion is such that the acceleration of the car is zero when it is moving with speed \(4U\).

  1. Given that the resistance is proportional to the car's speed, show that the distance \(X_1\) travelled by the car while it accelerates from speed \(U\) to speed \(2U\), is given by \[ \lambda X_1 = 2\ln \tfrac 9 5 - 1 \,, \] where \(\lambda= P/(16mU^3)\).
  2. Given instead that the resistance is proportional to the square of the car's speed, show that the distance \(X_2\) travelled by the car while it accelerates from speed \(U\) to speed \(2U\) is given by \[ \lambda X_2 = \tfrac43 \ln \tfrac 98 \,. \]
  3. Given that \(3.17<\ln 24 < 3.18\) and \(1.60<\ln 5 < 1.61\), determine which is the larger of \(X_1\) and \(X_2\).

Solution:

  1. \(\,\) \begin{align*} && F_{res} &= kv \\ && P &= Fv \\ v = 4U: && 0 &= F-F_{res} \\ \Rightarrow && 0 &= \frac{P}{4U} - 4Uk \\ \Rightarrow && k &= \frac{P}{16U^2} \\ \\ &&m v \frac{\d v}{\d x}&= \frac{P}{v} - \frac{P}{16U^2}v \\ \Rightarrow && X_1 &= \int_{v=U}^{v=2U} \frac{16U^2mv^2}{P(16U^2-v^2)} \d v \\ v = Ut&& &= \frac{16mU^2}{P} \int_{t=1}^{t=2}\left ( \frac{t^2}{16-t^2} \right)U\d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 + \frac{16}{16-t^2} \right) \d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 +\frac{2}{4+t} +\frac{2}{4-t} \right) \d t \\ &&&= \frac{1}{\lambda}\left (-1 + 2\ln(6)-2\ln(2)-2\ln(5)+2\ln(3) \right) \\ \Rightarrow && \lambda X_1 &= 2\ln \tfrac95-1 \end{align*}
  2. \(\,\) \begin{align*} && F_{res} = kv^2 \\ v = 4U: && 0 &= \frac{P}{4U} - 16U^2k \\ \Rightarrow && k &= \frac{P}{64U^3} \\ \\ && mv \frac{\d v}{\d x} &= \frac{P}{v} - \frac{P}{64U^3}v^2 \\ \Rightarrow && X_2 &= \int_{v=U}^{v=2U} \frac{64U^3mv^2}{P(64U^3-v^3)} \d v \\ &&&= \frac{64U^3m}{P} \int_{v=U}^{v=2U} \frac{v^2}{64U^3-v^3} \d v\\ v = Ut &&&= \frac{64U^3m}{P} \int_{t=1}^{t=2} \frac{U^2t^2}{64U^3-U^3v^3} U \d t\\ &&&= \frac{4}{\lambda} \int_1^2 \frac{t^2}{64-t^3} \d t \\ &&&= \frac{4}{\lambda} \left [ -\frac13\ln(64-t^3) \right]_1^2 \\ &&&= \frac{4}{3\lambda} \ln (63/56) \\ \Rightarrow && \lambda X_2 &= \tfrac43 \ln \tfrac98 \end{align*}
  3. \(\,\) \begin{align*} && 2\ln \tfrac95 - 1 &\overset{?}{>} \frac43 \ln \frac98 \\ \Leftrightarrow && 4 \ln 3 - 2\ln 5 - 1 &\overset{?}{>} \frac83\ln 3 -4 \ln 2 \\ \Leftrightarrow && \frac43(3\ln 3 + 3\ln 2 - 2 \ln 3) &\overset{?}{>} 2 \ln 5 + 1\\ \Leftrightarrow && \frac43\ln 24 &\overset{?}{>} 2 \ln 5 + 1\\ \end{align*} The \(LHS\) is \(>4.22\). The \(RHS\) is \(< 4.22\), and therefore our inequality holds, in particular, \(X_1 > X_2\).

View