SOLUTION: A car leaves town X for town Y 120 km away at an average speed of 80 km/hr at 8:30 a.m. At the same time a bus leaves town Y for town X at an average speed of 60 km/hr. At 8:30 a.m

Question 1172824: A car leaves town X for town Y 120 km away at an average speed of 80 km/hr at 8:30 a.m. At the same time a bus leaves town Y for town X at an average speed of 60 km/hr. At 8:30 a.m, a cyclist leaves town Y for town X at an average speed of 30 km/hr
(a)Calculate the time when the bus meets the car to the nearest minute. (3mks)
(b)Calculate the distance between the car and the bus by the time the cyclist meets the car (4marks)
(c)If the bus upon reaching town X stops for 10 minutes then starts its journey back to Y. Calculate how far from X the bus meets the cyclist
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
distance=speed*time
the car and the bus are approaching each other at 140 km/h.
That means that they will cover the 120 km distance in 120/140 hour or 6/7 hr. This is 360/7 min or 51 minutes to the nearest minute.
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The cyclist meets the car: combined closing speed is 110 km/h and will meet at 120/110 min after 8:30. They meet in 12/11 hour.
In that time the car has gone 80.273 km. towards Y and is 39.727 km from Y.
the bus has gone 60*12/11 or 60.454 km. towards X and is 59.546 km from Y.
The absolute value of those differences, or 19.819 km, is the distance they are apart.
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The bus has been gone 2 hours + 10 minutes and is now starting from X.
In that 2h10m, the cyclist has gone 30*2/1/6=65 km
So the two are closing at 90 km/h and meet in 65/90 hr. or 13/18 hr.
The bus has gone 60*13/18 km or 43.33 km from X
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