Question

a train
cover a distance of 90 km
unorm speed. Had the speed been 15 km/hour toe.
ve taken 30 minutes less for jou
for journey. Find the original speed of the train​

Answer 1

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Answer 2

Answer:

Original Speed = 12.50 (approx)

Step-by-step explanation:

Here,

s = 90km = 90,000 m

New speed, v1 = (v + 4.17) m/s [15 km/h = 4.17 m/s and let v = original speed]

New time, t1 = (t - 1800) sec [t = original time]

Now,

s = v1 * t1

=> s = (v+4.17) * (t -1800)

=> s = vt - 1800v + 4.17t - (4.17*1800)

=> s = s - 1800v + 4.17 * s/v - (4.17*1800)

=> 1800v - 4.17s/v + (4.17*1800) = 0

=> 1800v^2 - 4.17 s + (4.17*1800)v = 0 [multiplying both side by v]

=> v^2 - (50*4.17) + 4.17 v = 0

Solving the equation, we get

v = 12.50