Question

$OA$ and $OB$ are two roads enclosing an angle of $120^\circ. X$ and $Y$ start from $'O'$ at the same time. $X$ travels along $OA$ with a speed of 4 km/hour and $Y$ travels along $ OB$ with a speed of $3$ km/hour. The rate at which the shortest distance between $X$ and $Y$ is increasing after $1$ hour is

• A. $37\, km/hour$
• B. $\sqrt {13} \,km /hour$
• C. $\sqrt {37}\, km /hour$
• D. $13\, km/hour$

Answer 1

Answer: (C)

Given, speed of x is $4 \,km/ h$ and y is 3 km/ h.
After time t the distance covered by x is 4t and y is 3t.

Let shortest distance between x and y = A.
Then by cosine law
$A^2 = (4 t )^2 + (3 t)^2 - (4 t )(3 t )2 \; \cos \; 120^{\circ}$
$ \Rightarrow A^{2} = 16t^{2} + 9t^{2} -24 t^{2} \left(-\frac{1}{2}\right)$
$ \Rightarrow A^{2} = 25t^{2} + 12t^{2} $
$\Rightarrow A^{2} = 37 t^{2} \,\,\,\,\dots(i)$
$ \Rightarrow A = \sqrt{37 t} $
If $t = 1 h,$ then $A = \sqrt{37}$ km
Now, differentiating Eq. (i) w.r.t. t, we get
2AA' = 37 (2t)
After t = 1 h, we get
$2 \sqrt{37} A' = 2(37)$
$\Rightarrow \; A '= \sqrt{37}$
Thus, rate at which shortest distance A change with time is $\sqrt{37}$ km/h.