A train, travelling at a uniform speed for \(360\,\text{km}\), would have taken \(48\) minutes less to travel the same distance if its speed were \(5\,\text{km/h}\) more. Find the original speed of the train.
45 km/h
Step 1: Assume the original speed
Let the original speed of the train be \(v\,\text{km/h}\).
Step 2: Write the time formula
Time taken to travel a distance = \(\dfrac{\text{distance}}{\text{speed}}\).
Step 3: Time at the original speed
Time = \(\dfrac{360}{v}\) hours.
Step 4: Time at the increased speed
If the speed is increased by 5 km/h, new speed = \(v + 5\).
Time = \(\dfrac{360}{v+5}\) hours.
Step 5: Difference in times
It is given that the difference in times is 48 minutes.
Convert 48 minutes to hours: \(48\,\text{minutes} = \dfrac{48}{60} = \dfrac{4}{5}\,\text{hours}\).
Step 6: Form the equation
\(\dfrac{360}{v} - \dfrac{360}{v+5} = \dfrac{4}{5}\).
Step 7: Simplify
Take LCM of denominators: \(v(v+5)\).
\(\dfrac{360(v+5) - 360v}{v(v+5)} = \dfrac{4}{5}\).
\(\dfrac{1800}{v(v+5)} = \dfrac{4}{5}\).
Step 8: Cross multiply
\(1800 \times 5 = 4 \times v(v+5)\).
\(9000 = 4v^2 + 20v\).
Step 9: Simplify equation
Divide everything by 2: \(4500 = 2v^2 + 10v\).
Rearrange: \(2v^2 + 10v - 4500 = 0\).
Divide by 2: \(v^2 + 5v - 2250 = 0\).
Step 10: Solve quadratic equation
Factorise: \(v^2 + 5v - 2250 = 0\).
\((v + 50)(v - 45) = 0\).
So, \(v = -50\) or \(v = 45\).
Step 11: Choose positive value
Since speed cannot be negative, the answer is \(v = 45\).
Final Answer: The original speed of the train is 45 km/h.