Question:
The speed of a train changes from 36 km/h to 72 km/h in 10 sec. Calculate the distance travelled during this time.
Answer:
The initial speed = u = 36 Km/hr
Final speed = 72 Km/hr
Time taken t = 20 sec
We need the distance travelled which can be got from the third equation of motion v2 = u2 + 2as
We know a = v-u /t = 72 – 36 / (20 * 1/60 min * 1/60 hr)
= 36 * 60 * 60 / 20 = 6480 Km/hr2
Hence, from v2 = u2 + 2as , we get s = (v2 - u2) / 2a
s = (v2 - u2) / 2a = (72)2 – (36)2 / 2 * 6480 = 0.3 Km
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