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June 2019 p42 q6
3422
A car has mass 1000 kg. When the car is travelling at a steady speed of \(v \text{ m s}^{-1}\), where \(v > 2\), the resistance to motion of the car is \((Av + B) \text{ N}\), where \(A\) and \(B\) are constants. The car can travel along a horizontal road at a steady speed of \(18 \text{ m s}^{-1}\) when its engine is working at \(36 \text{ kW}\). The car can travel up a hill inclined at an angle of \(\theta\) to the horizontal, where \(\sin \theta = 0.05\), at a steady speed of \(12 \text{ m s}^{-1}\) when its engine is working at \(21 \text{ kW}\). Find \(A\) and \(B\).
Solution
First, calculate the driving force (DF) for each case using the formula \(DF = \frac{P}{v}\), where \(P\) is the power and \(v\) is the speed.
Case 1: Horizontal road
\(DF = \frac{36000}{18} = 2000 \text{ N}\)
The resistance force is \(Av + B\), so:
\(18A + B = 2000\)
Case 2: Inclined road
\(DF = \frac{21000}{12} = 1750 \text{ N}\)
The resistance force is \(Av + B\) plus the component of weight along the incline: