Model 3 Train Vs Bridge/Platform Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Using Rule 10,

Length of train = x metre (let)

Speed of train

= $\text"Length of train and bridge"/ \text"Time taken in crossing"$

${x + 500}/100 = {x + 250}/60$

${x + 500}/5 = {x + 250}/3$

5x + 1250 = 3x + 1500

5x - 3x = 1500 - 1250

2x = 250 ⇒ x = $250/2$ = 125 metre

Answer: (a)

Speed of train

= $\text"Sum of length of both trains"/ \text"Time taken"$

${60 × 5}/18 = {110 + 170}/t = 280/t$

t = ${280 × 18}/{60 × 5}$ = 16.8 seconds.

Answer: (c)

Using Rule 10,

Speed of train = 90 kmph

= $90 × 5/18$ = 25 m/sec

Distance covered

= 230 + 120 = 350 m

Time taken = $350/25$

= 14 seconds

Answer: (b)

Speed of train = 36kmph

= $36 × 5/18$ = 10 m/sec.

If the length of bridge be x metre, then

10 = ${200 + x}/55$

200 + x = 550

x = 550 - 200 = 350 metre.

Using Rule 10,

Here, x = 200 m

u = 36 km/hr, ${36 × 5}/18$ m/second

= 10 m/sec

y = ?, t = 55 sec

using t = ${x + y}/u$

55 = ${200 + y}/10$

y = 550 - 200 = 350 m

Answer: (a)

Using Rule 10,

Speed of train

= $\text"Length of platform and train"/ \text"Time taken incrossing"$

= $({100 + 50}/10)$ metre/second

= 15 metre/second

= $(15 × 18/5)$ kmph = 54 kmph

Answer: (d)

Speed of train

= $\text"length of platform and train"/ \text"Time taken in crossing"$

=$({450 +150}/20)$ m/sec.

= $(600/20)$ m/sec.

= $(30 × 18/5)$ kmph = 108 kmph.