Understanding the Problem: Cycling Across a Square Field

The question asks for the time it takes to cycle across a square field along its diagonal. We are given the area of the square field and the constant speed of the cyclist. To find the time, we need to determine the distance (the length of the diagonal) and use the relationship between distance, speed, and time.

Step-by-Step Solution to Find the Time

Here's how we can solve this problem:

1. Calculate the side length of the square from its area.
2. Calculate the length of the diagonal of the square using the side length.
3. Convert the speed from kilometers per hour to a suitable unit, such as meters per minute, to match the distance unit (meters) and the desired output unit (minutes).
4. Use the formula Time = Distance / Speed to find the time taken.

1. Calculating the Side Length of the Square

The area of a square is given by the formula: Area = side × side = \(s^2\). We are given the area is 7200 m\(^2\).

Let \(s\) be the side length of the square field.

Area = \(s^2\)

\(7200 \text{ m}^2 = s^2\)

To find \(s\), we take the square root of the area:

\(s = \sqrt{7200}\)

We can simplify the square root:

\(s = \sqrt{3600 \times 2} = \sqrt{3600} \times \sqrt{2} = 60\sqrt{2} \text{ m}\)

So, the side length of the square field is \(60\sqrt{2}\) meters.

2. Calculating the Length of the Diagonal

The diagonal of a square with side length \(s\) can be found using the Pythagorean theorem or the formula \(d = s\sqrt{2}\). The diagonal is the hypotenuse of a right-angled triangle formed by two sides of the square.

Diagonal \(d = s\sqrt{2}\)

Substitute the value of \(s\) we found:

\(d = (60\sqrt{2}) \times \sqrt{2}\)

\(d = 60 \times (\sqrt{2} \times \sqrt{2})\)

\(d = 60 \times 2\)

\(d = 120 \text{ m}\)

The distance the cyclist needs to cover is the length of the diagonal, which is 120 meters.

3. Converting the Speed

The cyclist's speed is given as 4 km/h. The distance is in meters, and we want the time in minutes (as per the options). We should convert the speed to meters per minute.

• 1 kilometer = 1000 meters
• 1 hour = 60 minutes

Speed = 4 km/h

Speed = \(\frac{4 \text{ km}}{1 \text{ h}} = \frac{4 \times 1000 \text{ m}}{60 \text{ min}}\)

Speed = \(\frac{4000 \text{ m}}{60 \text{ min}}\)

Simplify the fraction:

Speed = \(\frac{400}{6} \text{ m/min} = \frac{200}{3} \text{ m/min}\)

The cyclist's speed is \(\frac{200}{3}\) meters per minute.

4. Calculating the Time Taken

Now we can calculate the time using the formula:

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

Distance = 120 m

Speed = \(\frac{200}{3}\) m/min

Time = \(\frac{120 \text{ m}}{\frac{200}{3} \text{ m/min}}\)

To divide by a fraction, we multiply by its reciprocal:

Time = \(120 \times \frac{3}{200} \text{ minutes}\)

Time = \(\frac{120 \times 3}{200} \text{ minutes}\)

Cancel out common factors (e.g., 10 from numerator and denominator, then 2):

Time = \(\frac{12 \times 3}{20} \text{ minutes}\)

Time = \(\frac{3 \times 3}{5} \text{ minutes}\) (dividing 12 and 20 by 4)

Time = \(\frac{9}{5} \text{ minutes}\)

The time taken to cross the field diagonally is \(\frac{9}{5}\) minutes.

Comparing this result with the given options, we find that it matches option 2.

Revision Table: Key Concepts

Additional Information: Unit Conversions and Formulas

It's crucial to pay attention to units when solving physics or measurement problems. Mixing units like kilometers and meters or hours and minutes will lead to incorrect answers.

• To convert kilometers to meters, multiply by 1000. (1 km = 1000 m)
• To convert hours to minutes, multiply by 60. (1 h = 60 min)
• To convert hours to seconds, multiply by 3600. (1 h = 3600 s)
• To convert speed from km/h to m/s, multiply by \(\frac{1000}{3600}\) or \(\frac{5}{18}\).
• To convert speed from km/h to m/min, multiply by \(\frac{1000}{60}\) or \(\frac{50}{3}\).

The formulas used are fundamental:

• Area of a square: \(A = s^2\)
• Diagonal of a square: \(d = s\sqrt{2}\) (derived from \(s^2 + s^2 = d^2\))
• Relationship between Distance (D), Speed (V), and Time (T): \(D = V \times T\), \(V = \frac{D}{T}\), \(T = \frac{D}{V}\)

Understanding these basic geometric and kinematic relationships is key to solving such problems efficiently.