Problems on Trains Questions Answers Formulas, Tips and Tricks

1. Problem on trains Formulae and Important facts

1. Conversion from km/hr to m/s

xkm/hr=(x∗518)m/sec$$\begin{array}{rl}x& km/hr=\left(x\ast \frac{5}{18}\right)m/sec\end{array}$$

2. Conversion from m/s to km/hr

xm/sec=(x∗185)km/h$$\begin{array}{rl}x& m/sec=\left(x\ast \frac{18}{5}\right)km/h\end{array}$$

3. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.

5. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.

6. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
The time taken by the trains to cross each other

(a+b)(u+v)sec$$\begin{array}{r}\frac{(a+b)}{(u+v)}sec\end{array}$$

Please note this is just formula of time which is distance upon speed.
We are just adding two distances and two speeds

8. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:

The time taken by the faster train to cross the slower train =

(a+b)(u−v)sec$$\begin{array}{r}\frac{(a+b)}{(u-v)}sec\end{array}$$

Please note as trains are moving in the same directions so we used (u-v)

9. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A′sspeed):(B′sspeed)=(b√:a−−√)