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1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.2. A farmer moves along the boundary of a square field of side $ 10 \mathrm{~m} $ in $ 40 \mathrm{~s} $. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from its initial position?3. Which of the following is true for displacement?
(a) It can not be zero.
(b) Its magnitude is greater than the distance travelled by the object.
1. An object has moved through a distance. Yet it can have zero displacement, if it returns back to its starting point.
Displacement is the straight line distance between the initial point and the final point, or it is the shortest distance between the two points. For example, going on a lap around in a circular stadium returning to the starting point, the displacement is zero though the distance is a positive quantity.
2.
Given,
Length of the side of a square field = 10 m
Time taken to move along square field = 40 s
Total time taken by the farmer to walk around the square field = 2 min 20 sec. = (60 × 2 + 20) = 140 seconds. $\left(\because 1m=60sec\right)$
In 40 sec farmer covers 40 m (perimeter of the square = 4 × 10 m = 40 m)
Thus,
Distance covered in 40 s = 40 m
Distance covered in 1 s = $\frac {40}{40}m$
Distance covered in 140 s = $\frac {40}{40}\times 140\ m$ = 140 m
Therefore,
$Total\ round\ completed=\frac{Total\ distance\ covered}{Distance\ covered\ in\ 1\ round}$
$Total\ round\ completed=\frac{140}{40}$
$Total\ round\ completed= 3.5$
Hence the farmer covered 3.5 rounds.
Therefore, if the farmer starts moving from point A of the square field, then he reaches point C on covering 3.5 rounds.
Now, we know that,
Displacement = Shortest distance
= AC.
Here, ABC is a right-angled triangle.
Therefore, by Pythagoras theorem
Hypotenuse2 = Perpendicular2 + Base2
$A{C}^{2}=A{B}^{2}+B{C}^{2}$
$A{C}^{2}={10}^{2}+{10}^{2}$
$A{C}^{2}=100+100$
$A{C}^{2}=200$
$\sqrt{A{C}^{2}}=\sqrt{200}$ (applying root both side)
$AC=\sqrt{200}$
$AC=\sqrt{2\times 100}$
$AC=10\sqrt{2}\ m$
$AC=10\times 1.41$ $\left(\because \sqrt{2}=1.41\right)$
$AC=14.1\ m$
Thus, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is 14.1m.
3. (a) False - As displacement can be zero (when an object returns back to its initial point).
(b) False - As compared above the magnitude of the object is 1.41 m, whereas the distance travelled by the object is 140m (3.5 round), which is greater than the magnitude.
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