A train with a uniform speed passes a 122 meters long platform in 17 seconds and a 210 meters long bridge in 25 seconds. The speed of the train is:

Question

Prepp Team · Accepted Answer

Solving the Train Speed Problem

This problem involves a train moving at a uniform speed and passing objects of known lengths (a platform and a bridge). To solve this, we need to understand the distance covered by the train in each scenario.

When a train passes a platform or a bridge, the total distance covered by the train is equal to the length of the train plus the length of the platform or bridge.

Let's define the variables:

$L$ = Length of the train (in meters)
$v$ = Speed of the train (in meters per second, m/s)

We are given two scenarios:

The train passes a 122-meter platform in 17 seconds.
The train passes a 210-meter bridge in 25 seconds.

Using the formula: Distance = Speed $\times$ Time, we can set up two equations.

Setting Up Equations for Train Speed and Length

For the platform:

The distance covered is the length of the train plus the length of the platform.

Distance $= L + 122$ meters

Time $= 17$ seconds

Speed $= v$ m/s

So, the first equation is:

$L + 122 = v \times 17$

$L + 122 = 17v$

$L = 17v - 122$ (Equation 1)

For the bridge:

The distance covered is the length of the train plus the length of the bridge.

Distance $= L + 210$ meters

Time $= 25$ seconds

Speed $= v$ m/s

So, the second equation is:

$L + 210 = v \times 25$

$L + 210 = 25v$

$L = 25v - 210$ (Equation 2)

Solving for the Train's Speed

We now have two expressions for the length of the train, $L$. Since the length of the train is the same in both cases, we can equate Equation 1 and Equation 2:

$17v - 122 = 25v - 210$

Now, we solve this equation for $v$:

Add 210 to both sides:

$17v - 122 + 210 = 25v - 210 + 210$

$17v + 88 = 25v$

Subtract $17v$ from both sides:

$17v + 88 - 17v = 25v - 17v$

$88 = 8v$

Divide by 8:

$v = \frac{88}{8}$

$v = 11 \text{ m/s}$

The speed of the train is 11 meters per second.

Converting Speed to kmph

The question asks for the speed in kilometers per hour (kmph). We need to convert the speed from m/s to kmph. The conversion factor is:

$1 \text{ m/s} = \frac{18}{5} \text{ kmph}$

So, to convert 11 m/s to kmph, we multiply by $\frac{18}{5}$:

Speed in kmph $= 11 \times \frac{18}{5}$

Speed in kmph $= \frac{198}{5}$

Speed in kmph $= 39.6 \text{ kmph}

Therefore, the speed of the train is 39.6 kmph.

Scenario	Distance Covered	Time Taken	Equation
Passing Platform (122m)	Length of train + 122m	17 seconds	$L + 122 = v \times 17$
Passing Bridge (210m)	Length of train + 210m	25 seconds	$L + 210 = v \times 25$

Revision Table: Train Speed Calculation

Step	Description	Calculation
1	Define variables and set up equations based on Distance = Speed $\times$ Time for platform.	$L + 122 = 17v \implies L = 17v - 122$
2	Set up equations based on Distance = Speed $\times$ Time for bridge.	$L + 210 = 25v \implies L = 25v - 210$
3	Equate the expressions for train length $L$ and solve for speed $v$ in m/s.	$17v - 122 = 25v - 210 \implies 8v = 88 \implies v = 11$ m/s
4	Convert speed from m/s to kmph.	$11 \times \frac{18}{5} = \frac{198}{5} = 39.6$ kmph

Additional Information: Speed, Distance, and Time Concepts

Speed: Rate at which an object moves. Speed = Distance / Time.
Distance: The total path covered by the object. Distance = Speed $\times$ Time.
Time: Duration taken to cover the distance. Time = Distance / Speed.
Units Conversion: To convert speed from m/s to kmph, multiply by $\frac{18}{5}$. To convert speed from kmph to m/s, multiply by $\frac{5}{18}$.
Passing an Object (Platform/Bridge): When a train of length $L_{\text{train}}$ passes a stationary object of length $L_{\text{object}}$, the total distance covered by the train's front (or any point on the train) from the moment it touches the object to the moment it leaves the object is $L_{\text{train}} + L_{\text{object}}$.
Uniform Speed: Means the speed remains constant throughout the journey.