Cyclist A and B cycled atan average speed of 15 km/h and 20 km/h respectively from the same starting point X on the same route. Cyclist B started his journey 6 mins after cyclist A had started. i) What were the distances travelled by cyclist A and B 1 h cyclist A had started his journey from point X? ii) Dis cyclist B overtake cyclist A within the first hour of cyclist A's journey?

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.

answer; i believe the correct answer is (40,000 j) good luck

f=ma newton's second law 7x8=56n