Answer:
i. Cyclist A travelled 15 km 1 h cyclist A had started his journey from point X
Cyclist B travelled 18 km 1 h cyclist A had started his journey from point X
ii. cyclist B overtake cyclist A 6 km from the same starting point X.
Explanation:
From the question,
- Cyclist A and B cycled at an average speed of 15 km/h and 20 km/h respectively.
- Cyclist B started his journey 6 mins after cyclist A had started.
Let the cyclist A time be t.
Then, we can write that
For Cyclist A
Speed = 15 km/h
Time = t mins
For Cyclist B
Speed = 20 km/h
Time = (t - 6) mins
i) To determine the distances travelled by cyclist A and B 1h cyclist A had started his journey,
For Cyclist A
Speed = 15km/h
Time = 1h = 60 mins
From the formula
Speed = Distance / Time
Then,
Distance = Speed × Time
Putting the values into the equation,
Distance = 15km/h × 1h
Distance = 15 km
∴ Cyclist A travelled 15 km 1 h cyclist A had started his journey from point X
For cyclist B
Speed = 20km/h
Time = 1h - 6mins = 60mins - 6mins = 54mins = 54/60 hour = 0.9 h
Also, from
Distance = Speed × Time
Putting the values into the equation
Distance = 20km/h × 0.9h
Distance = 18 km
∴ Cyclist B travelled 18 km 1 h cyclist A had started his journey from point X
ii) To determine the distance cyclist B overtake cyclist A, that is, when the distance covered by cyclist A equals that covered by cyclist B.
First, we will determine the time at which the distances covered by both cyclists were equal.
From
For Cyclist A
Speed = 15 km/h
Time = t hour
Distance = Speed × Time
Distance = 15t km
For Cyclist B
Speed = 20 km/h
Time = (60t - 6) mins = (t - 0.1) hour
Distance = 20 × (t - 0.1) = (20t - 2) km
Equate the distances
15t = 20t - 2
15t - 20t = -2
-5t = -2
5t = 2
t = 2/5
t = 0.4 hour
Hence, cyclist B overtake cyclist A 0.4 hour after cyclist A had started.
For the distance cyclist B overtake cyclist A,
From
Distance = (20t - 2) km
Distance = (20×0.4 - 2) km
Distance = (8 - 2) km
Distance = 6 km
Hence, cyclist B overtake cyclist A 6km from the same starting point X.
