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June 2015 p43 q3
3449
A car of mass 860 kg travels along a straight horizontal road. The power provided by the car’s engine is \(P\) W and the resistance to the car’s motion is \(R\) N. The car passes through one point with speed 4.5 m s\(^{-1}\) and acceleration 4 m s\(^{-2}\). The car passes through another point with speed 22.5 m s\(^{-1}\) and acceleration 0.3 m s\(^{-2}\). Find the values of \(P\) and \(R\).
Solution
Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, and applying Newton's second law \(F = ma\), we have:
At the first point:
\(\frac{P}{4.5} - R = 860 \times 4\)
At the second point:
\(\frac{P}{22.5} - R = 860 \times 0.3\)
Subtract the second equation from the first to eliminate \(R\):
\(\frac{P}{4.5} - \frac{P}{22.5} = 860(4 - 0.3)\)
Solving for \(P\):
\(P = 17900\) W
Substitute \(P\) back into one of the original equations to find \(R\):