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The maximum and minimum distances of a comet ....

The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12}$ m and $8.0 \times 10^{10} $ m respectively. If the speed of the comet at the nearest point is $6 \times 10^4 $ m/s, the speed at the farthest point is:

(A) $3.0 \times 10^3 $ m/s
(B) $1.5 \times 10^3 $ m/s
(C) $4.5 \times 10^3 $ m/s
(D) $6.0 \times 10^3 $ m/s

Solution

We have, $v_1 r_1 = v_2 r_2 $

$\therefore v_1 \times 1.6 \times 10^{12} = 6 \times 10^4 \times 8.0 \times 10^ {10} $

$\Rightarrow v_1 = 3.0 \times 10^3 $ m/s

Answer: (A)

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