Question

A man reduces his speed from \(20 \mathrm{~km} / \mathrm{h}\) to \(18 \mathrm{~km} / \mathrm{h}\). So, he takes 10 minutes more than the normal time. What is the distance travelled by him? (a) \(30 \mathrm{~km}\) (b) \(25 \mathrm{~km}\) (c) \(50 \mathrm{~km}\) (d) \(36 \mathrm{~km}\)

Short answer

a) \(30\) km b) \(50\) km c) \(80\) km d) \(100\) km #Answer# a) \(30\) km

Step-by-step solution

Write down the given information and necessary formulas

We are given: - Initial speed: \(20\mathrm{~km/h}\) - Reduced speed: \(18\mathrm{~km/h}\) - Time difference: \(10\mathrm{~minutes}\) = \(10/60\mathrm{~hours} = 1/6\mathrm{~hours}\) The important formulas we will need are: 1. \(time = \frac{distance}{speed}\)

Set up an equation for the time difference

Since the man takes \(10\) minutes more (or \(1/6\) hours) to travel the same distance when he reduces his speed, we can write an equation representing the time difference: \(\frac{distance}{18} - \frac{distance}{20} = \frac{1}{6}\)

Solve the equation for distance

To solve this equation, we first need to find a common denominator for the two fractions: \(\frac{20*distance - 18*distance}{(20)(18)} = \frac{1}{6}\) Now we combine the numerators: \(\frac{2*distance}{360} = \frac{1}{6}\) Then, cross-multiply to solve for distance: \(2*distance = 360 * \frac{1}{6}\) Finally, divide both sides by 2: \(distance = \frac{360}{6*2} = 30 \mathrm{~km}\)

Check the answer

Since the answer we found is one of the options given, our solution is correct. The distance travelled by the man is \(\boxed{30\mathrm{~km}}\). The correct answer is (a).

Key concepts

Understanding Speed and Time

When we talk about speed and time, we are dealing with two fundamental aspects of motion. Speed indicates how fast something is moving. It is usually expressed in units like kilometers per hour (km/hr) or meters per second (m/s). On the other hand, time is the duration that something is moving or remains in motion.

The relationship between speed, time, and distance is expressed by the basic formula:

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

For example, if you know the speed of a vehicle and how long it has been traveling, you can determine the distance it covered. This foundational concept is a stepping stone to solving more complex problems regarding motion.

Exploring Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause them. When solving problems in kinematics, like the one in this exercise, we focus primarily on aspects such as speed, velocity, and time-distance relationships.

In our exercise, the kinematics concept is embodied in the scenario of a man adjusting his speed. Originally moving at a speed of 20 km/hr, he reduces it to 18 km/hr, affecting his travel time. By understanding changes in speed, we can analyze their direct impact on distance and time, and hence, calculate the exact values using the kinematic equations.

Mastering Problem-Solving Techniques

Problem-solving in math often requires a systematic approach. When faced with a problem, like the one in this exercise, the first step is to clearly jot down the known values and determine the unknown. This helps in structifying the problem.

• Identify what is given. Here, we know the two speeds and the time difference.
• Set up relevant equations involving known formulas.
• Perform algebraic manipulations to solve for the unknown.

We crafted our equation based on the time difference experienced at two different speeds. This orderly approach significantly simplifies complex problems and leads us to the right solutions.

Solving with Algebraic Equations

Algebraic equations are mathematical statements expressing the equality of two expressions, often to solve for an unknown variable. In our distance problem, we used algebra to equate the time difference caused by changing speeds.

We start with the expression:

• \( \frac{\text{distance}}{18} - \frac{\text{distance}}{20} = \frac{1}{6} \)

By identifying a common denominator and performing algebraic operations, we simplify and solve this equation. Such manipulations are the core of algebra, allowing us to work elegantly with numbers and symbols to find unknown values. This kind of practice hones essential skills needed for tackling various mathematical problems effectively.