Pada saat terbang pada ketinggian tertentu suhu di dalam pesawat adalah 21°C, sedangkan suhu di luar pesawat 34°C. Setiap naik 80 meter, suhu udara di luar pesawat akan turun 0,50°C. Jika ketinggian pesawat naik 2.400 meter, berapakah suhu udara di luar pesawat?

The train accelerates at the rate of 20 for some time, until it's just exactly time to put on the brakes, decelerate at the rate of 100, and come to a screeching stop after a total distance of exactly 2.7 km.
The speed it reaches while accelerating is exactly the speed it starts decelerating from.
Speed reached while accelerating = (acceleration-1) (Time-1) = .2 time-1
Speed started from to slow down = (acceleration-2) (Time-2) = 1 time-2
The speeds are equal. .2 time-1 = 1 time-2 time-1 = 5 x time-2
It spends 5 times as long speeding up as it spends slowing down.
The distance it covers speeding up = 1/2 A (5T)-squared = 0.1 x 25 T-squared = 2.5 T-squared.
The distance it covers slowing down = 1/2 A (T-squared) = 0.5 T-squared.
Total distance = 2,700 meters. (2.5 + .5) T-squared = 2,700 T-squared = 2700/3 = 900
T = 30 seconds
The train speeds up for 150 seconds, reaching a speed of 30 meters per sec and covering 2,250 meters. It then slows down for 30 seconds, covering 450 meters.
Total time = 180 sec = 3 minutes, minimum .
Observation: This solution is worth more than 5 points.

Answered by AL2006 | 2024-06-10

This can be done by hit and trial easily.
Considering train accelerate for 13000s and deaccelerate for 1000s so total distance covered is = 20cm/s * 13000s + 100cm/s*100s = (260000 + 10000 )cm = 270000cm = 2.7km
So total time taken is - 13100 seconds

Answered by Pulkit | 2024-06-10

The minimum time for the train to travel 2.7 km is 180 seconds, or 3 minutes. This includes 150 seconds of acceleration and 30 seconds of deceleration. The calculations involve kinematic equations for distance travelled during acceleration and deceleration phases.
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Answered by AL2006 | 2024-12-24

Jawaban:15°C ketika di ketinggian 2.400m.Penjelasan:dik:ketinggian: h1 (ketinggian akhir) 2.400m, h2 (kenaikan ketinggian perskala) 80msuhu: T1 (penurunan suhu setiap naik 80 meter) = 0,50°Cdit: T2? (suhu akhir)jawab:T2=h1/h2.T1 =2.400m/80m.0,50°C =30.0,50°C =15°Cperbedaannnya adalah 19°C dari 34°C-15°C

Answered by JavelinEradicator | 2025-07-03