Question

A train running at uniform speed passes a bridge 275m long in 15 seconds and another bridge 425m long in 21 seconds. Find the speed of the train in $\dfrac{km}{hr}$ .
(a)28 $\dfrac{km}{hr}$
(b)110 $\dfrac{km}{hr}$
(c)60 $\dfrac{km}{hr}$
(d)90 $\dfrac{km}{hr}$

Answer 1

Hint: By using the 2 conditions given, find the speed of the train. Consider length of train as another variable. Find the 2 variables by 2 equations as the distances are in meters, we will get speed in $\text{m}\cdot \text{se}{{\text{c}}^{-1}}$ . Convert it into kmph for the required answer. Let the distance travelled be “d” m in time “t” sec. Then speed “S” is
$S=\dfrac{d}{t}\text{ }\dfrac{m}{s}$

Complete step-by-step answer:
For converting into Kmph multiply it by $\dfrac{18}{5}$
$1\dfrac{m}{s}=\dfrac{3600}{1000}=\dfrac{18}{5}\text{ kmph}$

Let us assume the speed of the train to be as “S” m/sec.
Let us assume the length of the train to be as “L” m.
Given in question the train takes 15 seconds to cross the 275m bridge.
Given in question the train takes 21 seconds to cross the 425m bridge.
In each case the total distance travelled by train will be given by:
Distance travelled $=$ (Length of bridge) + (Length of train)
By basic knowledge of physics, we can say the speed formula as:
Speed of train $=\dfrac{\text{distance travelled}}{\text{time taken}}$
Case(i): The train took 15sec to cross the 275m bridge.
By the total distance formula given above we can say
Total distance $=275+l$
By speed formula given above, we can write “S” as:
$S=\dfrac{275+l}{15}\text{ }\dfrac{m}{s}$ ……………………………..(1)
Case(ii): The train took 21seonds to cross the 425m bridge.
By the total distance formula given above, we can say:
Total distance $=425+l$
By speed formula given above, we can write “S” as:
$S=\dfrac{425+l}{21}\text{ }\dfrac{m}{s}$ ………………………….(2)
By equating the equation (1) and equation (2), we get:
$\dfrac{275+l}{15}=\dfrac{425+l}{21}$
By cross multiplying the terms, we can write above equation:
$21\left( 275+l \right)=15\left( 425+l \right)$
By simplifying the above equation, we can write it as:
$5775+21l=6375+15l$
By grouping common terms together, we can write it as:
$6l=600$
By dividing with 6 on both sides, we get
$l=100$
By substituting ‘ l ’ in equation (1), we can write “S” as:
$S=\dfrac{100+275}{15}=25\dfrac{m}{s}$
Now, we have $v=25\dfrac{m}{s}$
We need in Kmph
$1\dfrac{m}{s}=\dfrac{3600}{1000}Kmph=\dfrac{18}{5}kmph$
By above we say
$25\dfrac{m}{s}=\dfrac{18}{5}\times 25=90kmph$
Option (d) is the correct answer.

Note: (1) Don’t forget to consider the length of the train to calculate the distance travelled.
(2) Convert $\dfrac{m}{s}$ into kmph carefully. Don’t multiply $\dfrac{5}{18}$ it should be $\dfrac{18}{5}$ .