Understanding Speed, Time, and Distance

This problem deals with the relationship between speed, time, and distance. When the distance covered is constant, speed and time are inversely proportional. This means if speed increases, time decreases, and if speed decreases, time increases.

The fundamental formula connecting these three is:

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

In this scenario, the man travels the same distance, but at a different speed, which results in a different time taken.

Setting Up the Problem Variables

Let's define the variables for the man's usual journey and the journey where he is late:

• Usual Speed = \(V\)
• Usual Time = \(T\)
• New Speed = \(V'\)
• New Time = \(T'\)
• Distance = \(D\)

According to the problem:

• The new speed is \(4/5\) of the usual speed: \( V' = \frac{4}{5}V \)
• The man is 15 minutes late, meaning his new time is 15 minutes more than his usual time: \( T' = T + 15 \text{ minutes} \)

Using the Distance Formula

Since the distance is the same for both journeys, we can set up an equation using the distance formula:

Distance (Usual) = Speed (Usual) \(\times\) Time (Usual)

\( D = V \times T \)

Distance (New) = Speed (New) \(\times\) Time (New)

\( D = V' \times T' \)

Because the distance is the same, we can equate the two expressions:

\( V \times T = V' \times T' \)

Solving for Usual Time (T)

Now, substitute the expressions for \(V'\) and \(T'\) into the equation:

\( V \times T = \left(\frac{4}{5}V\right) \times (T + 15) \)

Assuming the usual speed \(V\) is not zero (which it must be for travel), we can divide both sides of the equation by \(V\):

\( T = \frac{4}{5}(T + 15) \)

To eliminate the fraction, multiply both sides by 5:

\( 5 \times T = 5 \times \frac{4}{5}(T + 15) \)

\( 5T = 4(T + 15) \)

Now, distribute the 4 on the right side:

\( 5T = 4T + 4 \times 15 \)

\( 5T = 4T + 60 \)

Subtract \(4T\) from both sides to solve for \(T\):

\( 5T - 4T = 60 \)

\( T = 60 \)

The time \(T\) is in minutes because the delay (15 minutes) was given in minutes. So, the usual time is 60 minutes.

Converting to Hours

The options might be in hours. We know that 1 hour = 60 minutes.

\( T = 60 \text{ minutes} = 1 \text{ hour} \)

Summary of Times and Speeds

This confirms that if the usual time is 60 minutes, the new time is \(60 + 15 = 75\) minutes. Let's check if the speed ratio is correct: The time ratio is \(T / T' = 60 / 75 = 4/5\). Since speed is inversely proportional to time for the same distance, the speed ratio \(V' / V\) should be the inverse of the time ratio \(T / T'\). So, \(V' / V = T / T' = 4/5\), which matches the given information \(V' = \frac{4}{5}V\).

Therefore, the usual time taken to cover the same distance is 60 minutes or 1 hour.

Revision Table: Speed, Time, Distance

Here's a quick look at the core concepts used in this type of speed, time, and distance problem:

• Formula: \( \text{Distance} = \text{Speed} \times \text{Time} \)
• Relationship (Constant Distance): Speed is inversely proportional to Time (\( S \propto 1/T \)). This means \( S_1 T_1 = S_2 T_2 \) if distance is constant.
• Relationship (Constant Speed): Distance is directly proportional to Time (\( D \propto T \)). \( D_1/T_1 = D_2/T_2 \) if speed is constant.
• Relationship (Constant Time): Distance is directly proportional to Speed (\( D \propto S \)). \( D_1/S_1 = D_2/S_2 \) if time is constant.

Additional Information: Solving Speed Time Distance Problems

Speed, time, and distance problems often involve understanding the inverse relationship when distance is fixed. Here are some tips:

• Always ensure units are consistent (e.g., minutes or hours, km/h or m/s). Convert units if necessary.
• Clearly define your variables for usual and changed conditions.
• Use the constant factor (distance, speed, or time) to set up an equation.
• In problems where speed changes, leading to a change in time for the same distance, the ratio of speeds is the inverse of the ratio of times. If speed is \(S_1\) and \(S_2\), and times are \(T_1\) and \(T_2\), then \( S_1/S_2 = T_2/T_1 \).
• In this problem, the speed ratio is \(V' / V = (4/5)V / V = 4/5\). The time ratio should be the inverse, \(T / T' = 5/4\). We have \(T' = T + 15\). So, \(T / (T + 15) = 4/5\). Cross-multiply: \(5T = 4(T + 15)\), which gives \(5T = 4T + 60\), leading to \(T = 60\) minutes. This method using ratios is often quicker.