Standard +0.3 This is a straightforward application of the power-force-velocity relationship (P = Fv) combined with resolving forces on an incline. Students must equate driving force to resistance plus component of weight down the slope, then solve for v. While it involves multiple steps, each is standard and the question clearly signposts the method with no novel insight required.
7 In this question use \(g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
When a car, of mass 1200 kg , travels at a speed of \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) it experiences a total resistive force which can be modelled as being of magnitude \(36 v\) newtons.
The maximum power of the car is 90 kilowatts.
The car starts to descend a hill, inclined at \(5.2 ^ { \circ }\) to the horizontal, along a straight road.
Find the maximum speed of the car down this hill. [0pt]
[5 marks]
7 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
When a car, of mass 1200 kg , travels at a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ it experiences a total resistive force which can be modelled as being of magnitude $36 v$ newtons.\\
The maximum power of the car is 90 kilowatts.\\
The car starts to descend a hill, inclined at $5.2 ^ { \circ }$ to the horizontal, along a straight road.\\
Find the maximum speed of the car down this hill.\\[0pt]
[5 marks]\\
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q7 [5]}}