Question

During a rainy day, rain is falling vertically with a velocity

2 {m s}^{- 1}

. A boy at rest starts his motion with a constant acceleration of

2 m s^{- 2}

along a straight road. Find the rate at which the angle of the axis of an umbrella with vertical should be changed so that the rain always falls parallel to the axis of the umbrella.

Accepted Answer

Rain will appear to the boy as shown in the figure.

Velocity of rain with respect to boy,

${\vec{v}}_{rain,boy} = {\vec{v}}_{rain} - {\vec{v}}_{boy} = {\vec{v}}_{rain} + (- {\vec{v}}_{boy})$

At any time $t$ , the velocity of the boy is given by,

$v_{boy} = u + a t = a t (u = 0)$

Boy should hold his umbrella at an angle $θ$ from the vertical, then we can write as,

$\tan (θ) = \frac{v_{boy}}{v_{rain}} = \frac{a t}{v} . . . (1)$

Differentiating equation $(1)$ , we get:

$\frac{dtan (θ)}{d t} = \frac{d (\frac{a t}{v})}{d t}$

$\Rightarrow \sec^{2} θ \frac{d θ}{d t} = \frac{a}{v}$

$\Rightarrow \frac{d θ}{d t} = \frac{a}{v \sec^{2} (θ)} = \frac{a}{v [1 + \tan^{2} (θ)]}$

$\Rightarrow \frac{d θ}{d t} = \frac{a}{v [1 + \frac{a^{2} t^{2}}{v^{2}}]} = \frac{a}{[v^{2} + a^{2} t^{2}]}$

After plugging the value, we get:

Rate at which the angle of the axis of the umbrella with vertical is, $\frac{d θ}{d t} = \frac{2}{[4 + 4 t^{2}]} = \frac{1}{2 (1 + t^{2})}$ .

B M Sharma Solutions for Chapter: Kinematics II, Exercise 65: ILLUSTRATION 5.61

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 65: ILLUSTRATION 5.61

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 65: ILLUSTRATION 5.61 with Hints & Solutions

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