Standard +0.3 This is a straightforward application of P=Fv at maximum speed where driving force equals resistance. Students need to recognize that at maximum speed, acceleration is zero, so F=kv, then substitute into P=Fv to get k=P/v². The 'fully justify' requirement adds slight difficulty, but the setup is standard and requires only one key insight about equilibrium at maximum speed.
Kang is riding a motorbike along a straight, horizontal road.
The motorbike has a maximum power of 75 000 W
The maximum speed of the motorbike is \(50 \text{ m s}^{-1}\)
When the speed of the motorbike is \(v \text{ m s}^{-1}\), the resistance force is \(kv\) newtons.
Find the value of \(k\)
Fully justify your answer. [4 marks]
Question 5:
5 | Obtains 1500
PI by k = 30 | 3.4 | B1 | P = Fv
75000
F = =1500 N
50
Using Newton’s Second Law
F – kv = ma
At max speed a = 0
1500 – 50k = 0
k = 30
Obtains 1500−50k or
1500−kv or 1500=50k
Allow their 1500
PI by k = 30 | 3.4 | M1
Explains that at maximum speed
a = 0
or
Driving Force = Resistance
at maximum speed
OE | 2.4 | E1
Obtains k = 30 | 1.1b | A1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
Kang is riding a motorbike along a straight, horizontal road.
The motorbike has a maximum power of 75 000 W
The maximum speed of the motorbike is $50 \text{ m s}^{-1}$
When the speed of the motorbike is $v \text{ m s}^{-1}$, the resistance force is $kv$ newtons.
Find the value of $k$
Fully justify your answer. [4 marks]
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q5 [4]}}