Moderate -0.3 This is a standard M1 power-force-velocity problem requiring two simultaneous equations from P=Fv and F=ma+R. The setup is straightforward with clearly given data, though students must correctly apply P/v - R = ma at two different speeds and solve the resulting system—slightly easier than average due to its routine nature and clear structure.
2 A car of mass 1250 kg is travelling along a straight horizontal road with its engine working at a constant rate of \(P \mathrm {~W}\). The resistance to the car's motion is constant and equal to \(R \mathrm {~N}\). When the speed of the car is \(19 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) its acceleration is \(0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), and when the speed of the car is \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) its acceleration is \(0.16 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the values of \(P\) and \(R\).
ft wrong answer for R or P substituted into a correct linear equation
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For using $DF = P/v$ |
| | M1 | For using Newton's second law when $v = 19$ or when $v = 30$ |
| $P/19 - R = 1250 \times 0.6$ and $P/30 - R = 1250 \times 0.16$ | A1 | |
| $[19R + 19 \times 1250 \times 0.6 = 30R + 30 \times 1250 \times 0.16]$ | M1 | For attempting to eliminate P or R |
| $R = 750$ or $P = 28500$ | A1 | |
| $P = 28500$ or $R = 750$ | B1ft **[6]** | ft wrong answer for R or P substituted into a **correct** linear equation |
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2 A car of mass 1250 kg is travelling along a straight horizontal road with its engine working at a constant rate of $P \mathrm {~W}$. The resistance to the car's motion is constant and equal to $R \mathrm {~N}$. When the speed of the car is $19 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, and when the speed of the car is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $0.16 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the values of $P$ and $R$.
\hfill \mbox{\textit{CAIE M1 2011 Q2 [6]}}