Model 4 Train Vs Pole/Signal Post/Man Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Let the speed of train be x kmph

and its length be y km.

When the train crosses a man, it covers its own length

According to he question,

$y/{(x - 3) × 5/18}$ = 10

18y = 10 × 5(x –3)

18y = 50x –150 ..... (i)

and, $y/{(x - 5) × 5/18}$ = 11

18y = 55(x–5)

18y = 55x –275 .... (ii)

From equations (i) and (ii),

55x –275 = 50x–150

55x –50x = 275 - 150

5x = 125 ⇒ x = $125/5$ = 25

Speed of the train = 25 kmph

Here, $S_1 = 3, S_2 = 5, t_1 = 10/3600 , t_2 = 11/3600$

Speed of train = ${t_1S_1 - t_2S_2}/{t_1 - t_2}$

= ${{3 × 10}/3600 - {5 × 11}/3600}/{10/3600 - 11/3600}$

= ${-25}/3600 × 3600/{- 1}$ = 25 m/sec

Answer: (a)

Using Rule 1,

Speed of train = 36 kmph

= $({36 × 5}/18)$ m/sec = 10 m/sec.

Length of train = Speed × time

= 10 × 25 = 250 metre

Answer: (d)

Using Rule 6,

Let speed of train be x kmph

Relative speed = (x + 5) kmph

Length of train

= $100/1000$ km = $1/10$ km

${1/10}/{x + 5} = 36/{5 × 60 × 60}$

$1/{10(x + 5)} = 1/500$

x + 5 = 50 ⇒ x = 45 kmph

Answer: (c)

Using Rule 1,

Speed of train = 90 kmph

= $(90 × 5/18)$ m/sec. = 25 m/sec.

When a train crosses a post, it covers a distance equal to its own length.

Required time = $\text"Distance"/ \text"Speed"$

= $180/25$ = 7.2 seconds

Answer: (d)

Using Rule 1,

In crossing a man standing on platform, train crosses its own length.

Speed of train = $120/10 = 12$m/s

Answer: (b)

Speed of train = 50 kmph

= $({50 × 5}/18)$ m./sec. = $125/9$ m./sec.

Required time = $({100/125}/9)$ seconds

= $({100 × 9}/125)$ = 7.2 seconds