Question

An insect moves along a circular path of radius 10cm with constant speed. If it takes 1min to move from a point on the path to the diametrically opposite point, find a) the distance covered (b) the speed (c) the displacement (d)average velocity

Answer 1

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Answer 2

Answer:  

(a) The distance covered is 31.4 cm

(b) The speed is 0.0052 m/s

(c) The displacement is 20 cm

(d) The average velocity is 0.0033 m/s

Solution:

As given in question that it takes 1 min to move from one point to diametrically opposite point, that means in 1 min the insect will be completed half revolution. So, the distance covered will be equal to ‘half of the circumference’ of the circular path and displacement will be equal to the diameter of the circular path.

Thus,

$\bold{\text {Distance covered in 1 min} =\frac{\text {Circumference of one complete revolution}}{2}}$

$\text {Distance covered in 1 min} =\frac{2 \pi r}{2}=\pi r$

As the radius of the circular path is 10 cm, the distance covered in 1 min will be  

$\text {Distance}=\pi \times 10=3.14 \times 10=31.4\ \mathrm{cm}$

The displacement of half revolution in 1min will be equal to the diameter of circular path.

So ,

$\text {Displacement}=2 \times \text {radius of the circular path}$

$\text {Displacement}=2 \times 10=20\ \mathrm{cm}$

The average speed is defined as the ratio of distance with time

$\text {Average speed}=\frac{\text { Distance }}{\text {Time}}=\frac{31.4 \mathrm{cm}}{1 \mathrm{min}}$

$\text {Average speed}=\frac{\frac{31.4}{100} m}{60 s}=\frac{31.4}{6000}=0.0052\ \mathrm{m} / \mathrm{s}$

The average velocity is defined as the ‘rate of change’ of displacement with respect to time.

$\text {Average Velocity}=\frac{\text {Displacement}}{\text {Time}}=\frac{20 \mathrm{cm}}{1 \min }$

$\text {Average Velocity}=\frac{\frac{20}{100} m}{60 s}=\frac{20}{6000}\bold{=0.0033\ \mathrm{m} / \mathrm{s}}$