A train travels at a constant speed and crosses two persons moving in

1.	A train travels at a constant speed and crosses two persons moving in the same direction in 10.8 seconds and 11.4 seconds respectively. The first one was running at the speed of 4.8 km/h while the second one was running at the speed of 6.6 km/h. What is the speed of the train in km/h?1. 39.62. 40.23. 38.44. 39.0
Answer» Correct Answer - Option 4 : 39.0 Given: Time taken to cross first-person = 10.8 sec Time taken to cross second person = 11.4 sec Speed of first-person = 4.8 km/h Speed of second person = 6.6 km/h Concept used: When two objects are traveling in the same direction the relative speed will be the difference in their speeds (faster – slower) Speed = Distance/Time Calculation: let the speed of the train be x km/h Relative speed in first case = (x – 4.8) km/h \(distance = (x - 4.8)\times \frac{10.8}{3600}\) Relative speed in second case = (x – 6.6) km/h \(distance =(x-6.6 )\times\frac{11.4}{3600}\) Distance in both cases will be equal i.e. length of the train \((x-4.8)\times\frac{10.8}{3600}=(x-6.6)\times\frac{11.4}{3600}\) ⇒ \(10.8 x -51.84=11.4x-75.24 \) ⇒ \(\\0.6x =23.4\) ⇒ \(\\x = 39km/h \) ∴ The speed of train is 39 km/hr

A train travels at a constant speed and crosses two persons moving in the same direction in 10.8 seconds and 11.4 seconds respectively. The first one was running at the speed of 4.8 km/h while the second one was running at the speed of 6.6 km/h. What is the speed of the train in km/h?1. 39.62. 40.23. 38.44. 39.0

Answer» Correct Answer - Option 4 : 39.0

Given:

Time taken to cross first-person = 10.8 sec

Time taken to cross second person = 11.4 sec

Speed of first-person = 4.8 km/h

Speed of second person = 6.6 km/h

Concept used:

When two objects are traveling in the same direction the relative speed will be the difference in their speeds (faster – slower)

Speed = Distance/Time

Calculation:

let the speed of the train be x km/h

Relative speed in first case = (x – 4.8) km/h

\(distance = (x - 4.8)\times \frac{10.8}{3600}\)

Relative speed in second case = (x – 6.6) km/h

\(distance =(x-6.6 )\times\frac{11.4}{3600}\)

Distance in both cases will be equal i.e. length of the train

\((x-4.8)\times\frac{10.8}{3600}=(x-6.6)\times\frac{11.4}{3600}\)

⇒ \(10.8 x -51.84=11.4x-75.24 \)

⇒ \(\\0.6x =23.4\)

⇒ \(\\x = 39km/h \)

∴ The speed of train is 39 km/hr