Solving the Car Speed Problem: Finding Usual Speed

This problem involves the relationship between distance, speed, and time. We are given the total distance, the fact that the car was late but reached on time, and the increase in speed required to do so. We need to find the car's usual speed.

Setting Up the Equations

Let's define the variables:

• Let the usual speed of the car be \(v\) km/hr.
• Let the scheduled time for the trip be \(t\) hours.
• The distance to the destination is 480 km.

The fundamental relationship is Distance = Speed × Time.

Scenario 1: Usual Trip

If the car travels at its usual speed \(v\) for the scheduled time \(t\), the distance covered is 480 km.

This gives us the equation:

\(480 = v \times t \quad (1)\)

From this, we can express the scheduled time \(t\) in terms of the usual speed \(v\):

\(t = \frac{480}{v} \quad (2)\)

Scenario 2: Late Start and Increased Speed

The car left 4 hours later than the scheduled time. To still reach the destination on time, the time taken for the trip must be 4 hours less than the scheduled time \(t\). So, the actual time taken for the trip is \(t - 4\) hours.

The car increased its usual speed by 10 km/hr. So, the increased speed is \(v + 10\) km/hr.

Using the Distance = Speed × Time relationship for this scenario:

\(480 = (v + 10) \times (t - 4) \quad (3)\)

Solving the System of Equations

Now we have a system of two equations with two variables, \(v\) and \(t\). We can substitute the expression for \(t\) from equation (2) into equation (3).

Substitute \(t = \frac{480}{v}\) into equation (3):

\(480 = (v + 10) \left(\frac{480}{v} - 4\right)\)

Now, let's simplify and solve for \(v\).

\(480 = (v + 10) \left(\frac{480 - 4v}{v}\right)\)

Multiply both sides by \(v\) to eliminate the denominator:

\(480v = (v + 10)(480 - 4v)\)

Expand the right side of the equation:

\(480v = v(480) + v(-4v) + 10(480) + 10(-4v)\)

\(480v = 480v - 4v^2 + 4800 - 40v\)

Move all terms to one side to form a quadratic equation:

\(0 = 480v - 4v^2 + 4800 - 40v - 480v\)

\(0 = -4v^2 - 40v + 4800\)

Divide the entire equation by -4 to simplify:

\(0 = v^2 + 10v - 1200\)

Finding the Usual Speed

We now need to solve the quadratic equation \(v^2 + 10v - 1200 = 0\) for \(v\). We can factor this equation. We look for two numbers that multiply to -1200 and add up to +10. These numbers are +40 and -30.

So, we can factor the equation as:

\((v + 40)(v - 30) = 0\)

This equation gives two possible solutions for \(v\):

• \(v + 40 = 0 \implies v = -40\)
• \(v - 30 = 0 \implies v = 30\)

Speed cannot be negative in this context, so \(v = -40\) km/hr is not a valid physical solution. Therefore, the usual speed of the car is 30 km/hr.

Verification

Let's check if a usual speed of 30 km/hr satisfies the conditions.

• Usual speed \(v = 30\) km/hr.
• Scheduled time \(t = \frac{480}{v} = \frac{480}{30} = 16\) hours.
• Car leaves 4 hours late, so actual travel time is \(16 - 4 = 12\) hours.
• Increased speed is \(v + 10 = 30 + 10 = 40\) km/hr.
• Distance covered with increased speed and reduced time = Speed × Time = \(40 \times 12 = 480\) km.

The calculated distance matches the given distance, so the usual speed of 30 km/hr is correct.

The usual speed of the car is 30 km/hr.

Revision Table: Car Speed Calculations

Additional Information: Distance, Speed, and Time Problems

Distance, speed, and time problems are common in quantitative aptitude. They typically involve scenarios where one of the variables (distance, speed, or time) changes, affecting the others. Key concepts to remember include:

• The relationship \( \text{Distance} = \text{Speed} \times \text{Time} \).
• Converting units if necessary (e.g., km/hr to m/s).
• Understanding relative speed when objects move towards or away from each other.
• Solving equations, which may sometimes lead to linear or quadratic equations.

Always define your variables clearly and set up equations based on the information given for different parts of the journey or different scenarios.