The length of path covered by a moving body between its initial and final positions is called the distance covered by the body. Distance is a scalar quantity, and its SI unit is meter. Distance always increases with time.

Draw a graph for the following. Is it a linear graph? Side of square (in cm) 2 3 3.5 5 6 Perimeter (in cm) 8 12 14 20 24

Let us draw the graph by taking side of a square ( c m ) on x -axis and perimeter ( c m ) on y -axis. Now, for a suitable scale, We take x -axis as 1 u n i t = 1 c m and y -axis as 1 unit = 4 cm . From the given data, the graph follows as below: Hence, it is a linear graph.

A body is said to be in motion when its position changes with time with respect to an observer. The state of motion (or rest) not only depends on the body but also depends on the state of the observer. Note: Rest and motion are relative terms.

An athlete completes one round of a circular track of diameter 200 m in 40 s . What will be the distance covered and the displacement at the end of 2 m i n 20 s ?

Given that diameter of the circular track, d = 200 m The radius of the track, r = d 2 = 100 m Time of running, t = 2 min 20 s = 140 s We know that circumference of a circular path is C = 2 π r = 2 π ( 100 ) = 200 π m And given that athlete completes one round in 40 s . In 40 s , athlete covers a distance of 200 π m . In unit time, the athlete will cover a distance = 200 π 40 m . The total distance covered by athlete in 140 s = 200 × 22 40 × 7 × 140 = 2200 m . When the athlete runs for 140 s , he completes 3 full round and a half-round around the circle. That means after 140 s he will be at a point which is diametrically opposite to the starting point. Therefore, the displacement will be equal to the diameter of the circular path which is 200 m .

An athlete completes one round of a circular track of diameter 49 m in 20 s . Calculate the distance covered and displacement at the end of 30 s .

Given that, Diameter of the track, D = 49 m Time taken for one cycle, T = 20 s Total time taken by the athlete, t = 30 s Total rounds completed in 30 s is, n = t T = 30 20 = 1 . 5 We know that the distance per one round is equal to the circumference of a circle which is equal to πD . Thus, the total distance travelled (let it be x ) in 30 s is, x = nπD = 3 πD 2 = 3 2 × 22 7 × 49 = 231 m After completing one and half round, the athlete will be at the exact opposite point of the circle. That is, the displacement will be equal to the diameter which is 49 m .

EASY

Earn 100

The numerical ratio of displacement to distance for a moving object is :

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Important Questions on Motion

MEDIUM

Science>Physics>Motion>Measuring the Rate of Motion

Abdul, while driving to school, computes the average speed for his trip to be

20 k m / h .

On his return trip along the same route, there is less traffic, and the average speed is

30 k m / h .

What the average speed for Abdul’s trip is?

EASY

Science>Physics>Motion>Measuring the Rate of Motion

Define distance.

HARD

Science>Physics>Motion>Measuring the Rate of Motion

Draw a graph for the following. Is it a linear graph?

Side of square (in cm)	$2$	$3$	$3.5$	$5$	$6$
Perimeter (in cm)	$8$	$12$	$14$	$20$	$24$

MEDIUM

Science>Physics>Motion>Measuring the Rate of Motion

If a body covers a distance $x$ with speed $v_{1}$ and then it covers another distance $x$ with speed $v_{2}$ , find the average speed for the whole journey.

EASY

Science>Physics>Motion>Measuring the Rate of Motion

Define motion.

EASY

Science>Physics>Motion>Measuring the Rate of Motion

What is the SI unit of distance?

EASY

Science>Physics>Motion>Measuring the Rate of Motion

An athlete completes one round of a circular track of diameter