Solving the Train Speed Problem
This problem involves calculating the original speed of a train given details about delays caused by an accident that reduced its speed.
Understanding the Variables
- Let the original speed of the train be s km/h.
- Let the total distance of the journey be D km.
- The reduced speed after the accident is $\frac{4}{5}s$ km/h.
- Delays are given in minutes and need conversion to hours.
Scenario 1: Accident After 30 km
The accident occurs after travelling 30 km. The remaining distance is $(D - 30)$ km.
The time taken for the remaining distance at the reduced speed is:
$ \text{Actual Time}_1 = \frac{D - 30}{\frac{4}{5}s} = \frac{5(D - 30)}{4s} \text{ hours} $The original time for this distance would have been:
$ \text{Original Time}_1 = \frac{D - 30}{s} \text{ hours} $The delay is 45 minutes, which is $\frac{45}{60} = \frac{3}{4}$ hours.
The delay is the difference between the actual time and the original time:
$ \text{Actual Time}_1 - \text{Original Time}_1 = \text{Delay}_1 $ $ \frac{5(D - 30)}{4s} - \frac{D - 30}{s} = \frac{3}{4} $Simplifying the left side:
$ \frac{5(D - 30) - 4(D - 30)}{4s} = \frac{3}{4} $ $ \frac{D - 30}{4s} = \frac{3}{4} $Multiplying both sides by 4:
$ \frac{D - 30}{s} = 3 $ $ D - 30 = 3s \quad \quad (1) $Scenario 2: Accident After 48 km
The accident occurs 18 km farther than in scenario 1, so after $30 + 18 = 48$ km. The remaining distance is $(D - 48)$ km.
The time taken for the remaining distance at the reduced speed is:
$ \text{Actual Time}_2 = \frac{D - 48}{\frac{4}{5}s} = \frac{5(D - 48)}{4s} \text{ hours} $The original time for this distance would have been:
$ \text{Original Time}_2 = \frac{D - 48}{s} \text{ hours} $The delay is 36 minutes, which is $\frac{36}{60} = \frac{3}{5}$ hours.
The delay is the difference:
$ \text{Actual Time}_2 - \text{Original Time}_2 = \text{Delay}_2 $ $ \frac{5(D - 48)}{4s} - \frac{D - 48}{s} = \frac{3}{5} $Simplifying the left side:
$ \frac{5(D - 48) - 4(D - 48)}{4s} = \frac{3}{5} $ $ \frac{D - 48}{4s} = \frac{3}{5} $Multiply both sides by 4:
$ \frac{D - 48}{s} = \frac{12}{5} $ $ D - 48 = \frac{12s}{5} \quad \quad (2) $Calculating the Original Speed
Now we have a system of two linear equations with two variables, D and s:
- $D - 30 = 3s \implies D = 3s + 30$
- $D - 48 = \frac{12s}{5}$
Substitute the expression for D from equation (1) into equation (2):
$ (3s + 30) - 48 = \frac{12s}{5} $ $ 3s - 18 = \frac{12s}{5} $Multiply the entire equation by 5 to eliminate the fraction:
$ 5(3s - 18) = 12s $ $ 15s - 90 = 12s $Rearrange the terms to solve for s:
$ 15s - 12s = 90 $ $ 3s = 90 $ $ s = \frac{90}{3} $ $ s = 30 \text{ km/h} $The original speed of the train was 30 km/h.
Similar Questions
- A train overtakes two persons walking along a railway track. The first one walks at 9 km/hr. The other one walks at 12.6 km/hr. The train needs 13.5 and 15 seconds, respectively, to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
- A train 180 m long is running at a speed of 90 km/h. How long will it take to pass a post?
- Two trains 136 m and 185 m in length respectively are running in opposite directions, one at a speed of 70 km/h and the other at a speed of 65 km/h. In what time will they be completely clear of each other from the moment they meet?
- Find the time taken by a 450 m long train travelling at the speed of 80 km/h to cross a platform of length 150 m.
- Two trains start from stations A and B and travel towards each other at the speeds of 60 km/h and 70 km/h respectively. At the time of their meeting, the second train has travelled 100 km more than the first. The distance between A and B is:
- A 300-m-long train crosses an electric pole in 5 s. Find the speed of the train.
- Simplify the following. $\sqrt{2 + \sqrt{2 + 2\cos 4\theta}}$
- A train is moving at a uniform speed of 75 km/h. Find the time required by the train to cover a distance of 250 km.
- A train that is 110 m in length is running at 90 km/h. How much time will the train take to cross a bridge that is 180 m in length?
- The ratio between the speeds of two trains is 7 : 5. If the second train runs 400 km in 4 h, then the speed of the first train is:
Important Questions from Problem on Trains
Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.
If the route between G and C is closed, which one of the following stations need not be passed through while travelling from H to C?
A daily train is to be introduced between station A and station B starting from each end at 6 AM and the journey is to be completed in 42 hours. What is the number of trains needed in order to maintain the Shuttle Service?
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:
How long does a train 153 meters long running at the rate of 90 kmph take to cross a bridge 622 meters in length?
A train passes a 360 metre long platform in 40 seconds and a man standing on the platform in 16 seconds. The speed of the train is:
