Maths

Question

A train leaves City A for City B at 60 km/h. At the same time, another train leaves City B for City A at 90 km/h. The distance between the two cities is 450 km.However, every 30 minutes, the first train increases its speed by 10 km/h, and the second train decreases its speed by 5 km/h.Question:After how many hours will the two trains meet?

Answer

handwritten step-by-step solution is attached in the image.final answer: the two trains meet after 39/14 hours

Answer

Answer

two trains depart simultaneously from opposite ends of a 450 km route. the twist: their speeds don't remain constant. train a accelerates every 30 minutes, while train b gradually slows. this creates a dynamic problem requiring interval-by-interval calculation.
train a
departs city a heading towards city b
initial speed: 60 km/h
increases by +10 km/h every 30 min

train b
departs city b heading towards city a
initial speed: 90 km/h
decreases by −5 km/h every 30 min
total distance
the two cities are 450 km apart
both trains depart at
exactly the same time goal: find when they meet***the first 30 minutes
both trains begin their journeys at their initial speeds. train a rolls out at 60 km/h whilst train b charges forward at 90 km/h. together, they close a significant portion of the gap in just the first half hour.
traina &mdash; 60 km/h
distance covered: 60 × 0.5 = 30 km
train b &mdash; 90 km/hdistance covered: 90 × 0.5 = 45 km
interval summary
combined this interval: 75 km     remaining distance: 375 kmthe next 30 minutes
train a has now accelerated to 70 km/h following its first speed increase, whilst train b has dropped to 85 km/h. the combined coverage continues to grow, but the remaining gap remains substantial at over 300 km.
train a speed update
60km/h+10km/h=70km/h distance this interval: 70 × 0.5 = 35 km
train b speed update
90km/h−5km/h=85km/h
distance this interval: 85 × 0.5 = 42.5 km
interval coverage
35km+42.5km=77.5km cumulative total: 152.5 km
remaining distance
450km−152.5km=327.5km still over two-thirds to go!the next 30 minutes
the dynamics begin to shift in interval 4. train a has now surpassed train b in speed &mdash; 90 km/h versus 75 km/h &mdash; as acceleration overtakes deceleration. the cumulative distance reaches 315 km, with 135 km still remaining.
speed crossover
train a now travels faster than train b for the first time, having started slower and steadily accelerated.
train a: 90 km/h (+30 from start) train b: 75 km/h (−15 from start)interval distances
traina:90×0.5=45kmtrainb:75×0.5=37.5km interval total: 82.5 km
cumulative: 315 km remaining: 135 km
the next 30 minutes
with train a now at 100 km/h and train b reduced to 70 km/h, the combined closing speed remains strong at 170 km/h. after this interval, a mere 50 km separates the two trains &mdash; the final stretch is imminent.
100 km/h-train a speed &mdash; fastest yet
70km/h - train b speed&mdash; steadily slowing85km interval cover  combined this interval
 remaining 50km left after 5 intervals
after five complete 30-minute intervals &mdash; a total of 2.5 hours elapsed &mdash; the cumulative distance covered reaches 400 km, leaving just 50 km between the two trains.approaching the meeting point
in the final partial interval, train a reaches 110 km/h and train b slows further to 65 km/h. their combined closing speed is 175 km/h. with only 50 km remaining, we can calculate precisely when they meet.
1-train a final speed : 100+10=110km/h 2-train b final speed:70−5=65km/h 
combined speed :110+65=175km/htime to cover 50 km:50÷175=≈0.2857hours

t6 = 50 km/175 km/h≈ 0.2857 hours

Answer

by half-hour intervals, total distance covered:
75 + 77.5 + 80 + 82.5 + 85 = 400 km after 2.5 hours.
remaining distance = 450 − 400 = 50 km.
next combined speed = 110 + 65 = 175 km/h.
time for remaining 50 km = 50/175 = 2/7 hour.
total = 2.5 + 2/7 = 39/14 hours ≈ 2.786 hours.they meet after 2.786 hours, approximately 2 hours 47 minutes.

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