Solving Distance, Speed, and Time Problems

This question involves two cars traveling the same distance but at different average speeds, resulting in different travel times. We are given the speeds of both cars and the difference in their travel times. We need to find the distance between the two points.

Understanding the Relationship: Distance, Speed, Time

The fundamental relationship between distance, speed, and time is:

\(\text{Distance} = \text{Speed} \times \text{Time}\)

This can be rearranged to find time or speed:

• \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
• \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)

In this problem, we are interested in finding the distance, and we can use the time formula to set up an equation based on the given information.

Setting up the Equations

Let:

• \(D\) be the distance between A and B in km.
• \(v_X\) be the average speed of car X, which is 50 km/hr.
• \(v_Y\) be the average speed of car Y, which is 75 km/hr.
• \(t_X\) be the time taken by car X to travel from A to B in hours.
• \(t_Y\) be the time taken by car Y to travel from A to B in hours.

Using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), we can write the expressions for \(t_X\) and \(t_Y\):

• For car X: \(t_X = \frac{D}{v_X} = \frac{D}{50}\)
• For car Y: \(t_Y = \frac{D}{v_Y} = \frac{D}{75}\)

The problem states that car X takes 2 hours more than car Y for the journey. This gives us a relationship between \(t_X\) and \(t_Y\):

\(t_X = t_Y + 2\)

Solving for the Distance

Now, we can substitute the expressions for \(t_X\) and \(t_Y\) into the equation \(t_X = t_Y + 2\):

\(\frac{D}{50} = \frac{D}{75} + 2\)

To solve for \(D\), we need to eliminate the denominators. The least common multiple (LCM) of 50 and 75 is 150. Multiply every term in the equation by 150:

\(150 \times \left(\frac{D}{50}\right) = 150 \times \left(\frac{D}{75}\right) + 150 \times 2\)

\(3D = 2D + 300\)

Now, subtract \(2D\) from both sides of the equation:

\(3D - 2D = 300\)

\(D = 300\)

So, the distance between A and B is 300 km.

Verification

Let's check if this distance satisfies the given condition:

• If \(D = 300\) km, time taken by car X is \(t_X = \frac{300 \text{ km}}{50 \text{ km/hr}} = 6\) hours.
• If \(D = 300\) km, time taken by car Y is \(t_Y = \frac{300 \text{ km}}{75 \text{ km/hr}} = 4\) hours.

The difference in time is \(t_X - t_Y = 6 \text{ hours} - 4 \text{ hours} = 2 \text{ hours}\). This matches the information given in the problem (X takes 2 hours more than Y). Therefore, the calculated distance is correct.

Summary of Calculation Steps

Revision Table: Distance, Speed, and Time

Here's a quick look at the key components involved in this type of problem:

Additional Information: Relative Speed and Time Difference

This problem can also be conceptualized in terms of relative speed, though the algebraic method is straightforward. The time difference between two objects covering the same distance depends on the difference in their speeds.

• When two objects cover the same distance \(D\), and their speeds are \(v_1\) and \(v_2\), the time taken is \(t_1 = D/v_1\) and \(t_2 = D/v_2\).
• The difference in time is \(|t_1 - t_2| = \left|\frac{D}{v_1} - \frac{D}{v_2}\right| = D \left|\frac{1}{v_1} - \frac{1}{v_2}\right|\).
• In our case, \(|t_X - t_Y| = 2\) hours, \(v_X = 50\) km/hr, \(v_Y = 75\) km/hr. Since \(v_Y > v_X\), \(t_Y < t_X\), so \(t_X - t_Y = 2\).
• \(D \left(\frac{1}{50} - \frac{1}{75}\right) = 2\)
• \(D \left(\frac{3}{150} - \frac{2}{150}\right) = 2\)
• \(D \left(\frac{1}{150}\right) = 2\)
• \(D = 2 \times 150 = 300\) km.

This confirms the result obtained through the direct substitution method. Understanding these relationships is crucial for solving various problems involving motion.