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in Physics by (34.9k points)
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A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground-based observer? 

(Hint: Assume north to be \(\hat i\) direction and vertically downward to be \(-\hat j\). Let the rain velocity vbe \(a\hat i+b\hat j\). The velocity of rain as observed by the girl is always vr - vgirl. Draw the vector diagram/s for the information given and find a and b. You may draw all vectors in the reference frame of ground-based observer.)

1 Answer

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Let vr= \(a\hat i+b\hat j \)

Velocity of girl =vg = (5 m/s)\(\hat i\)

Velocity of rain w.r.t. girl

vrg=vr−vg= \((a\hat i+b\hat j)-5\hat i\)

= (a – 5)\(\hat i\)+ b\(\hat j\)

Hence, a – 5= 0 ⇒ a = 5

Now, vg= (10\(\frac{m}{s}\) )\(\hat i\)

vrg= vr−vg

= \((a\hat i+b\hat j)-10\hat i\)

= (a – 10)\(\hat i+b \hat j\)

As, angle appear 45º,

∴ tan 45º =\(\frac{b}{a-10}\)= 1

⇒b = a – 10 = 5 – 10 = −5

Hence, velocity of rain = \(a\hat i+b\hat j\)

⇒v= \(5\hat i-5\hat j\)

Speed of rain

= \(\begin{vmatrix} v_r\end {vmatrix}\) = \(\sqrt{(5)^2+(-5)^2}\)

= \(\sqrt50=5\sqrt2\,m/s\).

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