Question

Two particles are moving along two long straight lines, in the same plane with same speed $20cm/s.$ The angle between the two lines is ${60}^o$ and their intersection point is $O$. At a certain moment, the two particles are located at distance $3m$ and $4m$ from $O$, and are moving towards $O$. Subsequently, the shortest distance between them will be

• A. $50cm$
• B. $40\sqrt{2}cm$
• C. $50\sqrt{2}cm$
• D. $50\sqrt{3}cm$

Answer 1

The correct option is D $50\sqrt{3}cm$

At time $t=0$,the first particle is $300cm$ from $O$ and moving toward $O$ at a velocity of $20cm/s.$
The distance of this particle to $O$ as a function of time is:
$A=300-20t$
At the time $t=0$, the second particle is $400cm$ from $O$ and moving toward $O$ at a velocity of $20m/s.$
The distance of this particle to $O$ as a function of time is:
$B=400-20t$
The distance between the $2$ particles can be found using the cosine rule
$C^2=A^2+B^2-2AB\cos {60}^0$
$C^2=(300-20t{)}^2+(400-20t{)}^2-2(300-20t)(400-20t)\cos {60}^0$
$C^2=(400t^2-12000t+90000)+(400t^2-16000t+160000)-(400t^2-14000t+120000)$
$C^2=400t^2-14000t+130000$
$C=\sqrt{(400t^2-14000t+130000)}$
Now one needs to find t such that $C$ will be minimized.
If C is minimized at a certain t, then also $C^2$ will be minimized at that t.
So instead of finding the first derivative of $C$ (which is a square root,
so a lot of work), one can simply find the first derivative of $C^2$ (a
quadratic expression, which is a lot less work)
$C^2=400t^2-14000t+130000$
$\frac{d{C}^2}{d{t}^2}=800t-14000$
$800t-14000=0$
$t=17.5s$
At time $t=17.5s$ the distance between the $2$ particles is minimized.
The distance at that time will be:
$C=\sqrt{(400\times {17.5}^2-14000\times 17.5+130000)}$
$C=\sqrt{7500}$
$\Rightarrow C=50\sqrt{3}$