Question

a motorcyclists is moving with a velocity of 40 m/s towards north when it started raining in the absence of wind. if the velocity of rain is 30 m/s then calculate the velocity of the rain and it's direction as observed by the motorcyclist
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Answer 1

Answer:

50 m/s towards south with vertical downward by the angle of   $tan^{-1}(\frac{4}{3} )$

Explanation:

Let the unit vector along north direction be $\hat{i}$.

$\therefore the unit vector along south direction will be - \hat{i}\\\\ The unit vector along upward direction be \hat{j}\\\\\therefore The unit vector along downward direction will be - \hat{j}\\\\\therefore The velocity of the motorcyclists will be 40\hat{i}\\\\ The velocity of the rain fall will be - 30\hat{j}\\\\ The relative velocity of the rainfall with respect to the motorcyclists will be -30\hat{j}-40\hat{i}\\\\ The value of the relative velocity = \sqrt{(-30)^2+(-40)^2}$

$\implies\sqrt{900+1600}=\sqrt{2500}=50m/s$

The direction of the velocity will be appeared to the motorcyclists to south with respect to vertical downward direction by an angle $\Theta$ where

$\Theta will be equal to tan^{-1}(\frac{4}{3})$

Answer 2

Question-

A motorcyclist is moving with a velocity of 40 m/s towards the north when it started raining in the absence of wind. If the velocity of rain is 30 m/s then calculate the velocity of the rain and its direction as observed by the motorcyclist.

Answer-

As there is no wind, we can take that the rain is falling perpendicular to the road.

The rain is falling at 30 m/s.

The motorcyclist is moving at 40 m/s toward the north.

Since the rain is falling perpendicular to the road, we can form a right triangle. Let the sides of the triangle be the speed. Then,

A² = B² + C²

⇒ A² = 40² + 30²

⇒ A² = 1600 + 900

⇒ A² = 2500

⇒ A = √2500

⇒ A = 50 m/s

The rain is moving with a velocity of 50 m/s. Its direction as observed by the motorcyclist is south.

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