The problem involves a train traveling 240 km at an initial speed and then increasing its speed by 20 km/h after replacing its engine, which reduces the travel time by 1 hour. We need to determine the initial speed of the train before replacing the engine.

Let the initial speed of the train be $v$ km/h.

• The travel time at the initial speed is given by:
  $t_1 = \frac{240}{v}$

After the speed increases by 20 km/h, the new speed is $v + 20$ km/h.

• The travel time at the new speed is:
  $t_2 = \frac{240}{v + 20}$

The problem states that the new travel time is 1 hour less than the original travel time:
$t_1 - t_2 = 1$

Substituting the expressions for $t_1$ and $t_2$:
$\frac{240}{v} - \frac{240}{v + 20} = 1$

Now, we solve this equation for $v$.

Step 1: Simplify the equation

Combine the fractions:
$\frac{240(v + 20) - 240v}{v(v + 20)} = 1$

Simplify the numerator:
$\frac{240v + 4800 - 240v}{v(v + 20)} = 1$

$\frac{4800}{v(v + 20)} = 1$

Step 2: Multiply through by $v(v + 20)$:

$4800 = v(v + 20)$

Expand the right-hand side:
$4800 = v^2 + 20v$

Rearrange into standard quadratic form:
$v^2 + 20v - 4800 = 0$

Step 3: Solve the quadratic equation

We solve $v^2 + 20v - 4800 = 0$ using the quadratic formula:
$v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 20$, and $c = -4800$. Substituting:
$v = \frac{-20 \pm \sqrt{20^2 - 4(1)(-4800)}}{2(1)}$ $v = \frac{-20 \pm \sqrt{400 + 19200}}{2}$ $v = \frac{-20 \pm \sqrt{19600}}{2}$ $v = \frac{-20 \pm 140}{2}$

Step 4: Calculate the two roots

$v = \frac{-20 + 140}{2} = \frac{120}{2} = 60$ $v = \frac{-20 - 140}{2} = \frac{-160}{2} = -80$

Since speed cannot be negative, $v = 60$ km/h.

Final Answer:

The initial speed of the train before the engine was replaced is 60 km/h.

Would you like a detailed explanation for the steps or verification?

Here are 5 related questions you might explore:

1. How would the solution change if the travel distance were different?
2. Can you graph the relationship between speed and time for this problem?
3. What happens if the time reduction is less than 1 hour?
4. How does the quadratic formula help solve real-world problems?
5. What are alternative ways to solve this problem without the quadratic formula?

Tip: Always check the units (e.g., km/h vs. hours) to ensure they are consistent when solving motion problems!