Solving Speed and Time Problems: A Detailed Example

In the realm of speed and time problems, understanding how to manipulate algebraic equations can be quite powerful. Letrsquo;s explore a specific example: a man who walks a certain distance at a specific speed, and then walks a shorter distance at a faster speed, with a total time given. By following a series of steps, we can determine the original speed. This article will guide through the detailed process and provide insights into solving related problems.

Problem Statement

A man walks 16 km at a certain speed and 12 km at 2 km/h faster. If his total time is 6 hours, what is the original speed?

Step-by-Step Solution

Let the original speed be x km/h.

The man walks 16 km at this speed, so the time taken to walk 16 km is:

Time1 (frac{16}{x})

He then walks 12 km at a speed of x 2 km/h, so the time taken to walk 12 km is:

Time2 (frac{12}{x 2})

The total time for the journey is 6 hours, so we can set up the equation:

(frac{16}{x} frac{12}{x 2} 6)

To solve this equation, we first eliminate the fractions by multiplying through by x(x 2):

(16(x 2) 12x 6x(x 2))

Expanding both sides:

(16x 32 12x 6x^2 12x)

Combining like terms:

(28x 32 6x^2 12x)

Rearranging the equation to set it to zero:

(6x^2 - 16x - 32 0)

Dividing the entire equation by 2 for simplicity:

(3x^2 - 8x - 16 0)

Using the quadratic formula to solve for x:

(x frac{-b pm sqrt{b^2 - 4ac}}{2a})

Where a 3, b -8, c -16:

(x frac{8 pm sqrt{(-8)^2 - 4 cdot 3 cdot (-16)}}{2 cdot 3})

Calculating the discriminant:

((-8)^2 - 4 cdot 3 cdot (-16) 64 192 256)

Now substituting back into the formula:

(x frac{8 pm 16}{6})

This gives us two possible solutions:

(x frac{24}{6} 4)

(x frac{-8}{6}) (not a valid solution since speed cannot be negative)

Thus, the original speed is:

(boxed{4 text{ km/h}})

Verification

Letrsquo;s verify the solution:

At 4 km/h, the time taken for the 16 km walk is:

(frac{16}{4} 4 text{ hours})

The speed for the second part of the journey is:

(4 2 6 text{ km/h})

The time taken for the 12 km walk at 6 km/h is:

(frac{12}{6} 2 text{ hours})

The total time taken is:

(4 2 6 text{ hours})

This confirms that the original speed is indeed 4 km/h.

Additional Examples

Now, letrsquo;s consider another example:

Let the speed be x km/h:

(frac{20}{x} - frac{10}{x 3} 12)

Multiplying through by x(x 3):

(20(x 3) - 1 12x(x 3))

Expanding and simplifying:

(2 60 - 1 12x^2 36x)

(1 60 12x^2 36x)

(12x^2 26x - 60 0)

(frac{12x^2 26x - 60}{2} 6x^2 13x - 30)

(6x^2 13x - 30 0)

Solving using the quadratic formula:

(x frac{-13 pm sqrt{169 720}}{12} frac{-13 pm 29}{12})

(x 2 text{ or } x -frac{25}{12})

The valid solution is:

(x 2 text{ km/h})

Verification:

(frac{20}{2} 10 text{ hours})

(frac{10}{5} 2 text{ hours})

Total time 12 hours.

Summary

By methodically solving the equations and verifying the solutions, we can confidently determine the original speed.

Key Takeaways:

Understanding the relationship between speed, distance, and time. Using algebraic methods to solve complex problems. Verifying solutions to ensure accuracy.