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Measurement of Time and Motion - Slow or Fast

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The motion of an object can be classified as slow or fast based on the distance it covers in a specific interval of time.

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Speed is defined as the total distance covered by an object divided by the total time taken. It is represented as v=dtv = \frac{d}{t}.

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If the speed of an object moving along a straight line keeps changing, its motion is said to be non-uniform motion.

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If an object moving along a straight line maintains a constant speed, its motion is called uniform motion. In this case, the average speed is the same as the actual speed.

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The basic unit of time is the second (s), and the basic unit of speed is m/sm/s. Other units include km/hkm/h and cm/scm/s.

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Periodic motion, such as that of a simple pendulum, is used to measure time. A simple pendulum consists of a small metallic ball called a 'bob' suspended from a rigid stand by a thread.

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One complete to-and-fro motion of the pendulum is called an oscillation. The time taken to complete one oscillation is called its time period.

📐Formulae

Speed=Total distance coveredTotal time takenSpeed = \frac{\text{Total distance covered}}{\text{Total time taken}}

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}

Time Period=Total Time TakenNumber of Oscillations\text{Time Period} = \frac{\text{Total Time Taken}}{\text{Number of Oscillations}}

1 km/h=1000 m3600 s=518 m/s1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}} = \frac{5}{18}\text{ m/s}

💡Examples

Problem 1:

A car travels a distance of 270 km270\text{ km} in 4.5 hours4.5\text{ hours}. Calculate its speed in km/hkm/h and m/sm/s.

Solution:

Given: Distance (dd) = 270 km270\text{ km}, Time (tt) = 4.5 hours4.5\text{ hours}.

Speed in km/hkm/h: Speed=2704.5=60 km/hSpeed = \frac{270}{4.5} = 60\text{ km/h}

To convert to m/sm/s: Speed=60×518=30018≈16.67 m/sSpeed = 60 \times \frac{5}{18} = \frac{300}{18} \approx 16.67\text{ m/s}

Explanation:

We first use the standard speed formula to find the value in km/hkm/h. Then, we multiply the result by 518\frac{5}{18} to convert it into the SI base units of m/sm/s.

Problem 2:

A simple pendulum takes 32 s32\text{ s} to complete 2020 oscillations. What is the time period of the pendulum?

Solution:

Given: Total time = 32 s32\text{ s}, Number of oscillations = 2020.

Time Period=Total time takenNumber of oscillationsTime\ Period = \frac{\text{Total time taken}}{\text{Number of oscillations}} Time Period=3220=1.6 sTime\ Period = \frac{32}{20} = 1.6\text{ s}

Explanation:

The time period is the duration of exactly one oscillation. Dividing the total time by the total number of oscillations gives the time for a single cycle.

Problem 3:

A cheetah runs at a speed of 120 km/h120\text{ km/h}. How much distance will it cover in 10 minutes10\text{ minutes}?

Solution:

Given: Speed = 120 km/h120\text{ km/h}, Time = 10 minutes10\text{ minutes}. First, convert time to hours: 10 minutes=1060=16 hours10\text{ minutes} = \frac{10}{60} = \frac{1}{6}\text{ hours}

Now, calculate distance: Distance=Speed×TimeDistance = Speed \times Time Distance=120×16=20 kmDistance = 120 \times \frac{1}{6} = 20\text{ km}

Explanation:

To find the distance, the units of speed and time must be compatible. We convert minutes to hours before multiplying by the speed in km/hkm/h.