Question

Two people leave the same city to different destinations at the same time. Person $A$ leaves due east at a constant speed of $3\sqrt{2}$ kmph and person $B$ leaves due south east at a constant speed of $6$ kmph. How fast is the distance between them incresing $4$ hours later?

• A. $12\sqrt{2}$
• B. $6$
• C. $3\sqrt{2}$
• D. $24$

Answer 1

The correct option is C $3\sqrt{2}$

Let $O$ be the starting point

Let $x=$ Distance covered by person $A$
$y=$ Distance covered by person $B$
$z=$ Distance between persons $A$ and $B$
After $4$ hours, $x=3\sqrt{2}\times 4=12\sqrt{2}\text{km}$
$y=6\times 4=24\text{km}$
By Cosine Rule: $z^2=x^2+y^2-2xy\cos \theta \dots \dots \left(1\right)$
$\theta ={45}^{\circ }\Rightarrow \cos \theta =\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$
Value of $z$ after 4 hours
$z_{4\text{hr}}=\sqrt{12\times 12\times 2+24\times 24-2\times 12\times \sqrt{2}\times 24\times \frac{1}{\sqrt{2}}}$
$z_{4\text{hr}}=12\sqrt{2}\text{km}$
Let $\frac{dz}{dt}=z{}^{\prime },\frac{dx}{dt}=x{}^{\prime },\frac{dy}{dt}=y{}^{\prime }$
Differentiating equation $\left(1\right)$ w.r.t time
$2zz{}^{\prime }=2xx{}^{\prime }+2yy{}^{\prime }-2\cos \theta (xy{}^{\prime }+yx{}^{\prime })$
$\Rightarrow 12\sqrt{2}z{}^{\prime }=12\sqrt{2}\times 3\sqrt{2}+24\times 6-\frac{1}{\sqrt{2}}(12\sqrt{2}\times 6+24\times 3\sqrt{2})$
$\Rightarrow z{}^{\prime }=3\sqrt{2}\text{Kmph}$