# Motion - Class 9 - Science

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## Motion - Questions and Answers | NCERT | Science | Class 9

An athlete completes one round of a circular track of diameter $200 \mathrm{~m}$ in $40 \mathrm{~s}$. What will be the distance covered and the displacement at the end of 2 minutes $20 \mathrm{~s}$ ?

To find the distance covered and the displacement at the end of 2 minutes and 20 seconds by the athlete on a circular track with a diameter of $200 , \text{m}$, we need to calculate the following:

**Circumference of the track**: The circumference $ C $ of a circle with diameter $ d $ is given by the formula: $ C = \pi d $**Number of rounds in 2 minutes and 20 seconds**: First, convert the total time from minutes and seconds to seconds:

$$ 2 \text{ min} \times 60 \text{ sec/min} + 20 \text{ sec} = 120 \text{ sec} + 20 \text{ sec} = 140 \text{ sec} $$

Next, calculate the number of rounds $ N $ using the time taken for one round $ T $ (which is 40 seconds):

$$ N = \frac{\text{Total Time}}{T} $$

**Total distance covered**: The total distance $ D $ is the number of rounds multiplied by the circumference: $ D = N \times C $**Displacement at the end of 2 minutes and 20 seconds**: The displacement after completing full rounds is zero, as the athlete will be back at the starting point. If there is a fraction of the last round remaining, the displacement is the straight-line distance from the starting point to the point where the athlete stops. If the remaining distance corresponds to less than half a round, the displacement is the chord of the circle, which can be calculated using the angle traversed on the last incomplete round.

Let’s proceed with these calculations.

The circumference ( C ) of the circle is:

$$ C = \pi d = \pi \times 200 , \text{m} $$

The number of full rounds $ N $ completed in $ 140 , \text{sec} $ is:

$$ N = \frac{140}{40} $$

The number of full rounds is the integer part of $ N $, whereas any fractional part would represent an incomplete round.

Let's start by calculating the circumference $ C $ and the number of rounds $ N $.

The circumference ( C ) of the track is:

$$ C = \pi \cdot d = \pi \cdot 200 \text{ m} $$

Since $ \pi $ is approximately $ 3.14159 $, we have:

$$ C \approx 3.14159 \cdot 200 \text{ m} = 628.318 \text{ m} $$

Rounded to three significant figures, the circumference is about $ 628 \text{ m} $.

The number of rounds ( N ) is given by:

$$ N = \frac{140 \text{ s}}{40 \text{ s/round}} = 3.5 \text{ rounds} $$

The athlete completes $ 3 $ full rounds and half of another round in $ 140 \text{ s} $.

The distance covered, which is the total path length traveled, is: $ D = N \cdot C = 3.5 \cdot 628 \text{ m} \approx 2198 \text{ m} $

The displacement is the straight-line distance from the start to the end point. Since the athlete covers an exact number of full rounds plus a half round, the displacement is equal to the diameter of the circular track, which is $ 200 \text{ m} $.

Therefore, the distance covered is approximately $ 2198 \text{ m} $, and the displacement is $ 200 \text{ m} $.

Joseph jogs from one end A to the other end B of a straight $300 \mathrm{~m}$ road in 2 minutes 30 seconds and then turns around and jogs $100 \mathrm{~m}$ back to point $\mathrm{C}$ in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from $\mathrm{A}$ to $\mathrm{B}$ and (b) from $\mathrm{A}$ to $\mathrm{C}$ ?

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An artificial satellite is moving in a circular orbit of radius $42250 \mathrm{~km}$. Calculate its speed if it takes 24 hours to revolve around the earth.