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June 2018 p43 q6
3428
A car of mass 1400 kg travelling at a speed of \(v \text{ m s}^{-1}\) experiences a resistive force of magnitude \(40v \text{ N}\). The greatest possible constant speed of the car along a straight level road is \(56 \text{ m s}^{-1}\).
Find, in kW, the greatest possible power of the car’s engine.
Find the greatest possible acceleration of the car at an instant when its speed on a straight level road is \(32 \text{ m s}^{-1}\).
The car travels down a hill inclined at an angle of \(\theta^\circ\) to the horizontal at a constant speed of \(50 \text{ m s}^{-1}\). The power of the car’s engine is \(60 \text{ kW}\). Find the value of \(\theta\).
Solution
(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the resistive force. At the greatest speed, \(v = 56 \text{ m s}^{-1}\), the resistive force \(F = 40 \times 56\). Therefore, \(P = 40 \times 56 \times 56 = 125440 \text{ W} = 125 \text{ kW}\).
(ii) The driving force \(F_d\) is \(\frac{125440}{32}\). Using Newton's second law, \(F_d - 40 \times 32 = 1400a\). Solving for \(a\), we get \(a = 1.89 \text{ m s}^{-2}\).
(iii) For the car on an incline, the equation is \(\frac{60000}{50} + 1400g \sin \theta - 40 \times 50 = 0\). Solving for \(\sin \theta\), we have \(\sin \theta = \frac{800}{14000}\). Thus, \(\theta = 3.3^\circ\).