Have given below solutions to the two questions on Speed Time Distance.
1. City A to City B is a downstream journey on a stream which flows at a speed of 5km/hr. Boats P and Q run a shuttle service between the two cities that are 300 kms apart. Boat P, which starts from City A has a still-water speed of 25km/hr, while boat Q, which starts from city B at the same time has a still-water speed of 15km/hr. When will the two boats meet for the first time? (this part is easy) When and where will they meet for the second time?
When boat P travels downstream, it will effectively have a speed to 30kmph. Likewise, Q will have an effective speed of 10kmph. The relative speed = 40kmph. So, the two boats will meet for the first time after 300/40 hours (Distance/relative speed) = 7.5 hours (Actually, for this part we do not need the speed of the stream)
The second part is more interesting, because the speed of the boats change when they change direction. Boat P is quicker, so it will reach the destination sooner. Boat P will reach City B in 10 hours (300/30). When boat P reaches city B, boat Q will be at a point 100kms from city B.
After 10 hours, both P and Q will be traveling upstream
P’s speed = 20km/hr
Q’s speed = 10 km/hr => Relative speed = 10kmph
Q is ahead of P by 100 kms
P will catch up with Q after 10 more hours (Relative Distance/relative speed – 100/10).
So, P and Q will meet after 20 hours at a point 200 kms from city B
Travels 200km at 40kmph
the next 200km @ 50kmph and
the final 200km @ 60kmph
So, Bus A will be at a distance of 200km from city M after 5 hours, and at a distance of 400km after 9 hours, and reach N after 12 hours and 20 mins
Travels at an overall average speed of 50kmph, so will take 12 hours for the entire trip
So, Bus B will travel
160kms in the first 4 hours
200 kms in the next 4
and 240 in the final 4
So, both buses cross each other when they are in their middle legs.
After 5 hours, bus A will be at a position 200kms from city M. At the same time, bus B will be at a distance 210kms from city N (4*40+50).
The distance between them will be 190kms (600-200-210). Relative speed = Sum of the the two speeds = 50+50 = 100 kmph.
Time taken = 190/100 = 1.9 hours. = 1 hour and 54 minutes. So, the two buses will meet after 6 hours and 54 minutes. Bus B will have travelled 210 + 95 = 305 kms. So, the two buses will meet at a point that is 305 kms from City N and 295 kms from city A.