Question

A 100 metres long train is moving at 50 km/hr. An another train 120 metres
long is moving in opposite direction and crosses each other in 6 seconds.
Find the speed of the second train ?​

Answer 1

Given that:

• Length of first train = 100 m
• Speed of first train = 50 km/hr

• Length of second train = 120 m
• Speed of second train be x km/hr.

• Total time taken = 6 seconds

To Find:

The speed of the second train.

We know that:

• Distance = Speed × Time

We have:

Distance = Length of first train + Length of second train

↠ Distance = (100 + 120) = 220 m

Speed = Speed of first train + Speed of second train

↠ Speed = (50 + x) km/hr

Converting speed into m/sec:

• 1 km/hr = 5/18 m/sec
• (50 + x) km/hr = (50 + x) × 5/18 m/sec

↠ Time = 6 sec

Finding the speed of second train:

↣ 220 = (50 + x) × 5/18 × 6

↣ 220 × 18 = (50 + x) × 5 × 6

↣ 3960 = 1500 + 30x

↣ 3960 - 1500 = 30x

↣ 2460 = 30x

↣ 2460/30 = x

↣ 82 = x

↣ x = 82

Hence,

• The speed of the second train is 82 km/hr.

Answer 2

Answer:

Given :-

• A 100 metres long train is moving at 50 km/hr. An another train 120 metres long is moving in opposite direction and crosses each other in 6 seconds.

To Find :-

• What is the speed of the second train.

Solution :-

Let,

$\mapsto$ The speed of the second train = y km/hr

First, we have to find the total distance :

$\implies \sf Total\: Distance =\: 100\: m + 120\: m$

$\implies \sf Total\: Distance =\: (100 + 120)m$

$\implies \sf\bold{\purple{Total\: Distance =\: 220\: m}}$

Now, we have to find the total speed :

$\implies \sf Total\: Speed =\: 50\: km/hr + y\: km/hr$

$\implies \sf Total\: Speed =\: (50 + y)km/hr$

$\implies \sf\bold{\purple{Total\: Speed =\: (50 + y)km/hr}}$

Now, we have to convert the speed km/hr into m/s :

$\implies \sf Speed =\: (50 + y)km/hr$

$\implies \sf Speed =\: (50 + y) \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/hr =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf\bold{\purple{Speed =\: (50 + y) \times \dfrac{5}{18}\: m/s}}$

According to the question,

$\longrightarrow \sf (50 + y) \times \dfrac{5}{18} \times 6 =\: 220$

$\longrightarrow \sf (50 + y) \times \dfrac{30}{18} =\: 220$

$\longrightarrow \sf (50 + y) =\: 220 \times \dfrac{18}{30}$

$\longrightarrow \sf (50 + y) =\: \dfrac{\cancel{3960}}{\cancel{30}}$

$\longrightarrow \sf (50 + y) =\: \dfrac{132}{1}$

$\longrightarrow \sf 50 + y =\: 132$

$\longrightarrow \sf y =\: 132 - 50$

$\longrightarrow \sf\bold{\red{y =\: 82\: km/hr}}$

$\therefore$ The speed of the second train is 82 km/hr .