Unit 8
9 marks · NormalThis question connects compound measures (speed) and analysis of function domains.
Anna cycles 18 km in 1 hour 12 minutes. Calculate her average speed in km/h.
Express Anna's speed from part (a) in metres per second (m/s). Give your answer correct to 2 decimal places.
Consider the function , where is the number of hours cycled.
i) Write down a suitable domain for in this context, explaining your reasoning.
Give the value of and interpret your answer in the context of the problem.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 15 km/h
(b) 4.17 m/s
(c) x > 0; time taken must be positive (cannot divide by zero); x is a positive real number
(d) g(1.5) = 12; This means if Anna cycles for 1.5 hours, her average speed is 12 km/h.
Common mistakes: Incorrect time conversion in part a; Incorrect conversion factor km/h to m/s; Misinterpretation of domain for function; Not linking value of g(x) to speed
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Converts 1 hour 12 minutes to hours or uses correct formula
A1 (a) · 1 mark · Calculates speed = 15 km/h
M2 (b) · 1 mark · Multiplies speed by correct conversion factor (1 km/h = 0.2778 m/s) (follow-through)
A2 (b) · 1 mark · Gives answer 4.17 m/s (rounded to 2 d.p.) (follow-through)
B1 (c) · 2 marks · States domain (x > 0) with correct context (positive time only, no zero)
M3 (d) · 1 mark · Substitutes x = 1.5 into g(x)
A3 (d) · 1 mark · g(1.5) = 12
B2 (d) · 1 mark · Interprets the meaning (average speed for 1.5 h is 12 km/h)
Worked solution
a) 1 hour 12 minutes = 1 + (12/60) = 1.2 hours. Speed = distance / time = 18 km / 1.2 h = 15 km/h.
b) 1 km/h = 1000/3600 = 5/18 = 0.2778 m/s. 15 km/h × 0.2778 = 4.167 ≈ 4.17 m/s (2 d.p.)
c) For g(x) = 18/x in this context, x is time in hours. Must have x > 0 (cannot be zero or negative hours).
d) g(1.5) = 18 ÷ 1.5 = 12. This represents an average speed of 12 km/h if Anna took 1.5 hours to cover 18 km.