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Motion and Time - Slow and Fast Motion

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 by comparing the distance covered by it in a given interval of time.

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An object that covers a larger distance in a given time is said to be faster, while an object that covers a smaller distance in the same time is said to be slower.

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The most convenient way to find out which of the two or more objects is moving faster is to compare the distance moved by them in a unit time. This quantity is called SpeedSpeed.

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Average Speed is defined as the total distance covered divided by the total time taken. In common usage, the term 'speed' usually refers to average speed.

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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 motionNon-uniform\ motion.

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If an object moving along a straight line maintains a constant speed, it is said to be in Uniform motionUniform\ motion.

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The basic unit of time is the secondsecond (ss) and the basic unit of speed is m/sm/s (metres per second).

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Speed can also be expressed in other units such as km/hkm/h (kilometres per hour) or m/minm/min (metres per minute).

📐Formulae

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

Distance=Speed×TimeDistance = Speed \times Time

Time=DistanceSpeedTime = \frac{Distance}{Speed}

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

💡Examples

Problem 1:

A car travels a distance of 200 km200\ km in 4 hours4\ hours. What is the speed of the car?

Solution:

Given: Distance d=200 kmd = 200\ km, Time t=4 ht = 4\ h. Using the formula Speed=DistanceTimeSpeed = \frac{Distance}{Time}, we get Speed=2004=50 km/hSpeed = \frac{200}{4} = 50\ km/h.

Explanation:

The speed of the car is calculated by dividing the total distance by the total time taken, resulting in 50 km/h50\ km/h.

Problem 2:

Comparison of two motions: Object A covers 10 m10\ m in 2 s2\ s, and Object B covers 15 m15\ m in 5 s5\ s. Which one is faster?

Solution:

Speed of Object A: vA=10 m2 s=5 m/sv_A = \frac{10\ m}{2\ s} = 5\ m/s. Speed of Object B: vB=15 m5 s=3 m/sv_B = \frac{15\ m}{5\ s} = 3\ m/s. Since 5 m/s>3 m/s5\ m/s > 3\ m/s, Object A is faster.

Explanation:

By calculating the speed (distance per unit time) for both objects, we can directly compare their rates of motion. Object A has a higher speed, thus it is faster.

Problem 3:

Convert a speed of 72 km/h72\ km/h into m/sm/s.

Solution:

To convert km/hkm/h to m/sm/s, multiply by 518\frac{5}{18}: 72×518=4×5=20 m/s72 \times \frac{5}{18} = 4 \times 5 = 20\ m/s.

Explanation:

Since 1 km=1000 m1\ km = 1000\ m and 1 hour=3600 s1\ hour = 3600\ s, the conversion factor is 10003600\frac{1000}{3600}, which simplifies to 518\frac{5}{18}.