Solving the Train Speed, Distance, and Time Problem

This problem involves calculating the usual speed of a train and then determining the time it takes to cover a different distance at that usual speed. We are given the distance covered at a uniform speed and how the time changes when the speed increases.

Setting Up the Equations

Let's define the variables:

• Let the usual speed of the train be \(v\) km/h.
• Let the usual time taken to cover 450 km be \(t\) hours.

The relationship between distance, speed, and time is: Distance = Speed \(\times\) Time.

For the initial journey:

\(450 = v \times t\)       (Equation 1)

The problem states that if the speed had been 5 km/h more, the train would have taken 1 hour less.

• New speed = \(v + 5\) km/h.
• New time = \(t - 1\) hours.
• The distance is still 450 km.

Using the distance, speed, time relationship for the new scenario:

\(450 = (v + 5) \times (t - 1)\)        (Equation 2)

Solving for Usual Speed (v)

From Equation 1, we can express \(t\) in terms of \(v\):

\(t = \frac{450}{v}\)

Now, substitute this expression for \(t\) into Equation 2:

\(450 = (v + 5) \times \left(\frac{450}{v} - 1\right)\)

Expand the right side of the equation:

\(450 = v \times \frac{450}{v} - v \times 1 + 5 \times \frac{450}{v} - 5 \times 1\)

\(450 = 450 - v + \frac{2250}{v} - 5\)

Subtract 450 from both sides:

\(0 = -v + \frac{2250}{v} - 5\)

To eliminate the fraction, multiply the entire equation by \(v\) (assuming \(v \neq 0\)):

\(0 \times v = (-v) \times v + \left(\frac{2250}{v}\right) \times v - 5 \times v\)

\(0 = -v^2 + 2250 - 5v\)

Rearrange the terms to form a standard quadratic equation:

\(v^2 + 5v - 2250 = 0\)

Solving the Quadratic Equation

We can solve this quadratic equation for \(v\) using the quadratic formula: \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our equation \(v^2 + 5v - 2250 = 0\), we have \(a = 1\), \(b = 5\), and \(c = -2250\).

Calculate the discriminant, \(\Delta = b^2 - 4ac\):

\(\Delta = (5)^2 - 4(1)(-2250)\)

\(\Delta = 25 + 9000\)

\(\Delta = 9025\)

Now find the square root of the discriminant:

\(\sqrt{\Delta} = \sqrt{9025} = 95\)

Now apply the quadratic formula:

\(v = \frac{-5 \pm 95}{2(1)}\)

\(v = \frac{-5 \pm 95}{2}\)

This gives two possible values for \(v\):

1. \(v_1 = \frac{-5 + 95}{2} = \frac{90}{2} = 45\)
2. \(v_2 = \frac{-5 - 95}{2} = \frac{-100}{2} = -50\)

Since speed cannot be negative, we reject \(v_2 = -50\).

Therefore, the usual speed of the train is 45 km/h.

Calculating Time for 315 km at Usual Speed

The question asks for the time it will take to cover 315 km at the usual speed (45 km/h).

Time = Distance / Speed

Time = \(\frac{315 \text{ km}}{45 \text{ km/h}}\)

Time = \(\frac{315}{45}\) hours

We can simplify the fraction:

Time = \(\frac{315 \div 5}{45 \div 5} = \frac{63}{9}\) hours

Time = \(7\) hours

So, the train will take 7 hours to cover 315 km at its usual speed.

Verification

Let's check if the usual speed of 45 km/h satisfies the original conditions:

• Usual speed = 45 km/h
• Usual time for 450 km = \(\frac{450}{45} = 10\) hours
• Increased speed = \(45 + 5 = 50\) km/h
• Time for 450 km at increased speed = \(\frac{450}{50} = 9\) hours

The new time (9 hours) is indeed 1 hour less than the usual time (10 hours). The calculated usual speed is correct.

The time to cover 315 km at 45 km/h is \(\frac{315}{45} = 7\) hours.

The final answer is 7 hours.

Revision Table: Train Speed Calculation

Additional Information: Solving Word Problems

Solving word problems like this requires careful translation of the given information into mathematical equations. Here are some tips:

• Read Carefully: Understand what is given and what needs to be found.
• Assign Variables: Use letters to represent the unknown quantities. Clearly state what each variable represents.
• Formulate Equations: Write down the relationships between the variables based on the information in the problem. Look for keywords that indicate mathematical operations (e.g., "more than," "less than," "rate," "total").
• Solve the Equations: Use algebraic techniques to solve the system of equations or the single equation you've formed.
• Check Your Answer: Plug your solution back into the original word problem to make sure it makes sense and satisfies all the conditions.

Problems involving speed, distance, and time often lead to algebraic equations, sometimes including quadratic equations, as seen in this example.