Question

a train travelling with a speed 60 km/h catches and other train travelling in the same direction and then leaves it 120m behind it in 18 sec. The speed of the train is

1. 26km/h
2. 35km/h
3. 36km/h
4. 63km/h​

Answer 1

Let speed of the 2nd train is S m/sec.

And,60 kmph = =60×518=503 m/sec.

As trains are traveling in same distance,

Then Relative distance, =60×518

=503503−S

=12018

⇒50−3S3=203

⇒50−3S=30∴S=10m/sec

Or, Speed of the 2nd train = 10×185 = 36 kmph.

Answer 2

Answer:Here is your answer.... And please mark as brainliest

Step-by-step explanation:

Let speed of the 2nd train is S m/sec.

And,

60 kmph = $$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}$$ m/sec.

As trains are traveling in same distance, Then Relative distance,

$$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$

Or, Speed of the 2nd train = $${\text{10}} \times \frac{{18}}{5}$$ = 36 kmph.

So therefore option 3. 36 km/h is the correct option