Peter and David live 36 km apart. They leave their homes at 1 PM riding bicycles towards each other. Peter averages 8 km/h and David travels 10 km/h. What time do they meet at?
2006-10-30 20:23:29 · 5 answers · asked by Arnav 3 in Science & Mathematics ➔ Mathematics
at 3 pm
after 2 hours
they can travel with a combined speed of 18 km/hr since they are walking towards each other
distance = time x speed
36 km = time x 18 km/hr
time = [36 km]/ 18 km/hr
time = 2 hrs
1 pm + 2 hours = 3 pm
2006-10-30 20:32:34 · answer #1 · answered by lazareh 2 · 2⤊ 0⤋
The rate at which the two get closer is the sum of their speed. This is 18 km/h. Since they start 36 km apart, and 36 / 18 = 2, it takes them 2 hours to meet. So, 3 PM.
You can double check by seeing how far each would get in two hours. Peter would travel 8 * 2 = 16 km, and David would travel 10 * 2 = 20 km. Together they would travel 16 + 20 = 36 km, so it works out.
3 PM.
2006-10-30 20:34:01 · answer #2 · answered by Kevin L 2 · 1⤊ 0⤋
Since they leave at the same time, you can also use their average speed x 2. Average speed = 9 km/h so they will move toward each other cut the distance by 18km/h. So 2 hours = 3 pm.
2006-10-30 20:36:21 · answer #3 · answered by LaxPlayer35 1 · 0⤊ 0⤋
we can stop one at there starting point by using the iestain thiyori.
then we can think david travels 18km/h to peters house.
so he has to travl 36km with 18km/h.
so the time to meet=36/18
=2h
so they will meet at 3pm.
2006-10-30 20:37:07 · answer #4 · answered by amila 1 · 0⤊ 0⤋
They are approaching each other at 10+8=18kph
36km/18km/hr=2hr They will meet at 3PM
2006-10-31 05:20:34 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋