Speed \(=\frac{\text { distance }}{\text { time }}\) Time \(=\frac{\text { distance }}{\text { speed }}=\frac{20 \mathrm{~km}}{60 \mathrm{~km} \mathrm{~h}^{-1}}=\frac{1}{3} \mathrm{~h}\) \(=\frac{1}{3} \times 60=20\) minutes

Short Answer

Expert verified
Answer: 20 minutes

Step by step solution

01

Write down the given values and the formula to solve for time

Given: Distance \(= 20\) km Speed \(= 60\,km/h\) Formula to solve time: Time \(= \frac{\text{distance}}{\text{speed}}\)
02

Plug the given values into the formula for time

Time \(= \frac{20\, km}{60\, km/h}\)
03

Simplify the expression

Time \(= \frac{1}{3}\,h\)
04

Convert hours to minutes

There are 60 minutes in 1 hour. To convert hours to minutes, multiply the number of hours by 60: Time \(= \frac{1}{3} \times 60=20\) minutes The time it takes to cover the distance of 20 km with a speed of 60 km/h is 20 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed Distance Time Formula
Understanding the relationship between speed, distance, and time is crucial in the realm of physics. The speed distance time formula is a fundamental concept that essentially explains how these three variables interact. The formula is stated as Speed = Distance / Time. To solve for one of the variables, we can rearrange the formula accordingly.

For instance, if we need to find the time taken to travel a certain distance at a given speed, we can rearrange the formula to Time = Distance / Speed. It's important to ensure the units for speed and distance are compatible, typically kilometers per hour (km/h) for speed and kilometers (km) for distance when dealing with metric units.

This basic formula is highly versatile and is a stepping stone to solving more complex problems in physics involving motion and time.
Units Conversion
Solving physics problems often involves converting units to maintain consistency and accuracy. In the above exercise, we used units conversion to find the time in minutes rather than hours. Units conversion is essential because physicists around the world use different measurement systems, and having a common understanding is crucial for collaboration and communication.

To convert from hours to minutes, you multiply the time in hours by 60, since there are 60 minutes in one hour. For example, \( \frac{1}{3} \) hours equals \( \frac{1}{3} \times 60 = 20 \) minutes. This step is imperative to ensure the correctness of the final answer and to display the time in a unit that is more comprehensible for daily use.
Solving Physics Problems
Tackling physics problems effectively requires a methodical approach. Start with understanding the problem, identifying what is given and what needs to be found. Next, represent this with appropriate formulas. For example, in our problem, we began with the given distance and speed to find the time. Always write down the known values and the formula:

Distance = 20 km, Speed = 60 km/h, Time = Distance / Speed.

Next, insert the values into the formula, simplifying and solving step by step. After finding the answer, reflect on whether the result makes sense and if the units are in the most practical form for interpretation. A systematic approach minimizes errors and enhances problem-solving skills.
Motion and Time
Motion and time are deeply intertwined in physics. Motion refers to the change in position of an object over time. Hence, understanding time and its measurement is critical in analyzing motion. In physics problems, especially those related to kinematics, we often measure time in seconds, minutes, or hours, depending on the context of the situation.

When solving for motion-related problems, it's essential to use the correct time unit to collaborate with other units like speed and distance effectively. Time's role in motion can be illustrated in graphical representations, like distance-time graphs, which provide a visual understanding of how an object moves over a period. The simplicity and elegance of the relationship among speed, distance, and time allow us to calculate any of these quantities if the other two are known.

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Most popular questions from this chapter

The different kinds of motion are: (i) Translatory motion: \(\mathrm{A}\) bus moving on a road, the motion of a rising balloon, the free fall of a stone under gravity, the motion of a cricket ball when it is hit by a batsman are examples of translatory motion. Translatory motion is further classified as rectilinear motion and curvilinear motion. When an object moves along a straight path, its motion is said to be rectilinear motion. The marching of soldiers on a straight road, the motion of a car on a straight road, the motion of carrom board coin are examples of rectilinear motion. When an object moves along curved path, its motion is called curvilinear motion. A bus moving on a fly-over bridge, a car taking a turn, a football kicked from the ground into air all have curvilinear motion. (ii) Rotatory motion: In this type of motion, the object rotates about a fixed axis. The motion of blades of a ceiling fan, the spin motion of a top, the motion of turbine, the motion of the earth around the sun are all circular or rotatory motion. In some cases, the rotatory and translatory motions take place simultaneously. When a bicycle moves, its wheels undergo translatory and rotatory motion. (iii) Oscillatory motion: A boy on a swing moves to-and-fro (back and forth). The motion such as above, where an object moves to-and-fro is called oscillatory motion. Other examples of oscillatory motion are the motion of the pendulum of a clock, the motion of a needle of a sewing machine, the motion of a piston of an engine etc.

The distance moved by motor cycle, \(\mathrm{d}=10267-10237\) \(=30 \mathrm{~km}\) The time taken, \(\mathrm{t}=9: 00 \mathrm{am}-8: 30 \mathrm{am}=30 \mathrm{~min}\) \(=\frac{1}{2} \mathrm{~h}\) The speed of motor cycles \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{30 \mathrm{~km}}{\frac{1}{2} \mathrm{~h}}=60 \mathrm{~km} \mathrm{~h}^{-1}\).

The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)

We know \(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow \mathrm{T} \Rightarrow \sqrt{\ell}\) \(\mathrm{T}^{2}=4 \pi^{2} \frac{\ell}{\mathrm{g}} \Rightarrow \mathrm{T}^{2} \alpha \ell\) From above relationships, it is clear that as the length of simple pendulum increases, time period also increases and vice versa.

An object moves to and fro produces oscillatory motion. If the object moves to and fro at faster rate, it is called vibratory motion.

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