
Raindrops are falling vertically with a velocity of $10\,m{s^{ - 1}}$ To a cyclist moving on a straight road the raindrops appear to be coming with a velocity of $20\,m{s^{ - 1}}$ The velocity of cyclist is:
A. $10\,m{s^{ - 1}}$
B. $10\sqrt 3 \,m{s^{ - 1}}$
C. $20\,m{s^{ - 1}}$
D. $20\sqrt 3 \,m{s^{ - 1}}$
Answer
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Hint: In order to solve this question, we will use the concept of relative velocity which is if two bodies are moving with different velocities in different direction then velocity of one body with respect to another is called relative velocity and its relative velocity with which other body appears to be moving.
Formula Used: If two bodies are moving perpendicular to each other then, relative velocity of one body with respect to another body is calculated as
$v = \sqrt {{v_A}^2 + {v_B}^2} $
where, $v$ is relative velocity and ${v_A},{v_B}$ are the velocity of two bodies A and B moving perpendicular to each other.
Complete step by step answer:
According to the question, we have given that raindrops are falling vertically with velocity let’s say ${v_{raindrop}} = 10m{s^{ - 1}}$ and let ${v_{cyclist}}$ denotes for the velocity of the cyclist then, both velocity vectors are perpendicular to each other because cyclist is moving horizontally on ground in straight line whereas raindrops are falling vertically so, relative velocity let’s say v is given in the question $v = 20\,m{s^{ - 1}}$.So, using the relative velocity formula we have,
$v = \sqrt {{v_{raindrop}}^2 + {v_{cyclist}}^2} $
On putting the values we get,
$20 = \sqrt {100 + {v_{cyclist}}^2} $
$\Rightarrow 400 - 100 = {v_{cyclist}}^2$
$\therefore {v_{cyclist}} = 10\sqrt 3 m{s^{ - 1}}$
So, the velocity of the cyclist will be ${v_{cyclist}} = 10\sqrt 3 \,m{s^{ - 1}}$.
Hence, the correct option is B.
Note:It should be remembered that, here raindrops are assumed to be falling perfectly in vertical straight lines which in reality is not possible due to air drag and cyclist is assumed to be moving in perfect straight line horizontally, is it were not the case then we need to calculate the components of velocity of each body in both directions and then follows the relative velocity concept.
Formula Used: If two bodies are moving perpendicular to each other then, relative velocity of one body with respect to another body is calculated as
$v = \sqrt {{v_A}^2 + {v_B}^2} $
where, $v$ is relative velocity and ${v_A},{v_B}$ are the velocity of two bodies A and B moving perpendicular to each other.
Complete step by step answer:
According to the question, we have given that raindrops are falling vertically with velocity let’s say ${v_{raindrop}} = 10m{s^{ - 1}}$ and let ${v_{cyclist}}$ denotes for the velocity of the cyclist then, both velocity vectors are perpendicular to each other because cyclist is moving horizontally on ground in straight line whereas raindrops are falling vertically so, relative velocity let’s say v is given in the question $v = 20\,m{s^{ - 1}}$.So, using the relative velocity formula we have,
$v = \sqrt {{v_{raindrop}}^2 + {v_{cyclist}}^2} $
On putting the values we get,
$20 = \sqrt {100 + {v_{cyclist}}^2} $
$\Rightarrow 400 - 100 = {v_{cyclist}}^2$
$\therefore {v_{cyclist}} = 10\sqrt 3 m{s^{ - 1}}$
So, the velocity of the cyclist will be ${v_{cyclist}} = 10\sqrt 3 \,m{s^{ - 1}}$.
Hence, the correct option is B.
Note:It should be remembered that, here raindrops are assumed to be falling perfectly in vertical straight lines which in reality is not possible due to air drag and cyclist is assumed to be moving in perfect straight line horizontally, is it were not the case then we need to calculate the components of velocity of each body in both directions and then follows the relative velocity concept.
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