Question

An ant is at the corner of a cubical room of side a. The ant can move with a constant speed of u. The minimum time taken by the ant to reach the farthest corner of the cube is
a)$\dfrac{3a}{u}$
b)$\dfrac{\sqrt{3}a}{u}$
c)$\dfrac{\sqrt{5}a}{u}$
d)$\dfrac{(\sqrt{2}+1)a}{u}$

Answer 1

Hint: In the question it is given that the ant moves with the constant speed. To determine the minimum time that an ant can take to travel from one corner to the farthest corner depends on the path taken. Hence using the relation between the speed, time taken and the distance covered, we can determine the smallest possible time for the smallest path that it can take.
Formula used:
$u=\dfrac{d}{t}$

Complete answer:
Let us say that the ant is initially at point E. hence the farthest point with respect to E is A. The shortest distance from E to point A is AE. But the ant cannot move along this line as there is no path through which an ant can travel. The ant can only move on the surfaces of the cube. The closest point to A is ‘P’.
The points P, E and H form a right angled triangle. Hence  according to Pythagoras theorem, the hypotenuse will be the shortest distance PE. The magnitude of PE from the theorem we get,

If ant is at E then minimum distance from E to A is PE+PA
$EA=EP+PA$
Case 1. To calculate Distance EP
In right angled triangle PEH,
$EP$ =$\sqrt{x^2+a^2}$
Case 2. To calculate Distance PA
In right angled triangle ABP
$PA$ =$\sqrt{a^2+(a-x)^2}$
Here, x is variable which will give us the minimum distance to be travelled
Now, $EA=EP+PA$
$EA$ = $\sqrt{x^2+a^2} + \sqrt{a^2+(a-x)^2}$ - (1)
To find the value of x, we have to differentiate EA w.r.t x and equate them to zero.
Differentiate w.r.t $x$
$\dfrac{dEA}{dx}$ = $\dfrac{2x}{2\sqrt{x^2+a^2}} + \dfrac{2(a-x)(-1)}{\sqrt{a^2+(a-x)^2}}$
$\dfrac{dEA}{dx}$ = $\dfrac{x}{\sqrt{a^2+x^2}} - \dfrac{(a-x)}{\sqrt{a^2+(a-x)^2}}$
$\dfrac{dEA}{dx}$ = $0$      ( For minimum value of $EA$)
$\dfrac{x^2}{a^2+x^2}$ = $\dfrac{(a-x)^2}{2a^2-2ax+x^2}$
$x^2(2a^2-2ax+x^2)$ = $(a^2-2ax+x^2)(a^2+x^2)$
$2a^2x^2-2ax^3+x^4$ = $a^4-2a^3x+a^2x^2+x^2a^2-2ax^3+x^4$
$\Rightarrow a^3(a-2x)$ = $0$
$a$ can’t be equal to zero
$\therefore x= \dfrac{a}{2}$
Now substitute the value of $ x = \dfrac{a}{2}$ in equation (1)
$EA$ = $\sqrt{x^2+a^2} + \sqrt{a^2+(a-x)^2}$
$EA$ = $\sqrt{x^2+\dfrac{a^2}{4}} + \sqrt{a^2+\dfrac{a^2}{4}}$
$EA$ = $\dfrac{\sqrt{5a^2}}{2} + \dfrac{\sqrt{5a^2}}{2} $
$\Rightarrow EA$ =$\dfrac{2\sqrt{5}a}{2}$
Minimum distance between E to A is, EA =$\dfrac{2\sqrt{5}a}{2}$
To calculate the the minimum time required to ant to reach E to A is
Minimum time = $\dfrac{ \text{Minimum}\, \text {distance}}{{\text{Constant} \, \text{speed}}}$
Minimum time = $\dfrac{\sqrt{5}a}{4}$
Hence the correct answer of the above question is option C.

Note:
It is to be noted that the ant can make its first move to any of the points of the corners of the cube. The distance covered will always be the same. The ant is moving with constant speed, hence the above algorithm is precise to determine the minimum distance.