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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A river is flowing with speed of 1km/hr.A swimmer wants to go to point C starting from point A. He swims with a speed of 5 km/hr at an angle $\theta$with respect to the river. If $AB=BC=400m$ Then

a)time taken by the man in 6 mins

b) time taken by the man in 8 mins

c)the value of $\theta=45 ^{\circ}$

d)the value of $\theta=30 ^{\circ}$

Can you answer this question?
 
 

1 Answer

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Answer: Time taken to get across is 6 mins.
Since $ AB=BC$
We use vector addition and write $5 \sin \theta=5\ cos \theta+1$
$\Rightarrow 5(\sin \theta-\cos \theta)=1\; \rightarrow $$\sin \theta-\cos \theta=\large\frac{1}{5} \rightarrow $$\theta=53^{\circ}$
$\therefore$ the two options with $\theta = 45$ and $\theta = 30$ are incorrect.
Let;s calculate the time taken to get across as $ t =\large\frac{AB}{5 \sin \theta}$
$\qquad=\large\frac{0.400}{5 \times \Large\frac{4}{5}}$$=\large\frac{1}{10}$$\;hrs$$=\large\frac{1}{10} $$\times 60\; min = 6\; min$
answered Jun 28, 2013 by meena.p
edited Aug 11, 2014 by balaji.thirumalai
 

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