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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ .

Solution

For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics.


$tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $

$\Rightarrow \frac {3-r}{4} = \frac {r}{2}$

$\Rightarrow r=1 $

So, $R \equiv ( - 1,0)$

Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$