Question

$\huge\red{\tt{QUESTION :-}}$

This question is from Physics bolo of class 11, CBSE board.
(It is a numerical) On a 60km track, a train travels the first 30km at a uniform speed of 360 km/hr. How fast much the train travel the next 30 km km so as to average 40 km/h for the entire trip?



#$\bold\pink{Answer}$$\bold\orange{with}$$\bold\green{Quality}$​

Answer 1

Given that, on a 60km track, a train travels the first 30km at a uniform speed of 360 km/hr.

(Distance = 30 km and speed = 360 km/hr)

Total distance covered is 60 km.

Time = Distance/Speed

Assume that the time taken by the train in first half is t1 and in second time is t2.

t2 = 30/360 = 1/12 hr

Similarly, in second half the train travels 30 km and it's average speed is 40 km/hr.

Assume that the speed in second half is x km/hr.

t2 = 30/x hr

Average speed is defined as the ratio of total distance covered with respect to total time taken i.e.

Average speed = (Total distance covered)/(Total time taken)

40 = (30 + 30)/(1/12 + 30/x)

40 = 60/[(x + 360)/12x]

40 = 60*12x/(x + 360)

40x + 14400 = 720x

680x = 14400

x = 21.18

Therefore, the speed of the train in second half is 21.18 km/hr.

Answer 2

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♻️Given:-♻️

Total Distance (d) = 60 km

Speed (v₁) during the 1st half journey = 30 km/h

Average Speed = 40 km/h

♻️To Calculate:-♻️

Speed (v₂) for the 2nd half = ?

♻️Formula To be used:-♻️

✍️Average Speed =Total distance/Total time

♻️Solution:-♻️

Average Speed = total distance/total time

Average Speed = Total Distance/ t₁ + t₂

Where t₁ = 30/30 = 1 h

$40 = \frac{60}{1 + \frac{30}{v} }$

40 (v + 30) = 60 v

v = 60 km/h

Hence, The answer is 60 km/h.

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