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June 2017 p43 q6
3435
A car of mass 1200 kg is travelling along a horizontal road.
(i) It is given that there is a constant resistance to motion.
(a) The engine of the car is working at 16 kW while the car is travelling at a constant speed of 40 m s-1. Find the resistance to motion.
(b) The power is now increased to 22.5 kW. Find the acceleration of the car at the instant it is travelling at a speed of 45 m s-1.
(ii) It is given instead that the resistance to motion of the car is (590 + 2v) N when the speed of the car is v m s-1. The car travels at a constant speed with the engine working at 16 kW. Find this speed.
Solution
(i)(a) Using the formula for power, \(P = Fv\), where \(P = 16000\) W and \(v = 40\) m/s, we have:
\(16000 = F \times 40\)
Solving for \(F\), we get \(F = 400\) N.
(i)(b) With the increased power, \(P = 22500\) W and \(v = 45\) m/s, we find the force:
\(22500 = F \times 45\)
\(F = 500\) N.
Using Newton's Second Law, \(F - ext{resistance} = ma\):
\(500 - 400 = 1200a\)
\(a = \frac{1}{12} = 0.0833 \text{ m/s}^2\)
(ii) Given \(16000 = (590 + 2v)v\), solve for \(v\):
\(16000 = 590v + 2v^2\)
\(2v^2 + 590v - 16000 = 0\)
Solving this quadratic equation gives \(v = 25 \text{ m/s}\).