Question

$OA$ and $OB$ are two roads enclosing an angle of $120^{\circ }.X$ and $Y$ start from ${}^{\prime }{O}^{\prime }$ 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. $37km/hour$
• B. $\sqrt{13}km/hour$
• C. $\sqrt{37}km/hour$
• D. $13km/hour$

Answer 1

Answer: (C)

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