A man running on a horizontal road has half the kinetic energy of a boy of half of his mass. When the man speeds up by m/s then his kinetic energy becomes equal to kinetic energy of the boy, the original speed of the man is

Let the man's mass and speed in this scenario be M Kg and x m/s, respectively. For the kinetic energy that is, use the straightforward formula $K.E=\frac{1}{2}m{v}^2$ Let the boy's speed be y m/s. Create the x-y connection using the limitations of the question.

Let the man's speed be x m/s.
And the average man weighs M kg.
Assumedly, a man's kinetic energy when jogging on a horizontal road is equal to half that of a youngster whose mass is equal to that of a man.

hence, the boy's mass $=\frac{M}{2}kg$.

Let the boy's speed be y m/s.
Knowing that kinetic energy is expressed as

$\Rightarrow$$K.E=\frac{1}{2}m{v}^2$

Where, K.E = kinetic energy
m = mass of the body
v = velocity of the body by which it travels.

Thus, the boy's kinetic energy ${(K.E)}_b=\frac{1}{2}\left(\frac{M}{2}\right)y^2$Joule

Additionally, the man's kinetic energy ${(K.E)}_m\frac{1}{2}M{x}^2$ Joule

The fact that a man has just half the kinetic energy of a youngster is now accepted as fact.

Therefore,

${(K.E)}_m=\frac{1}{2}{(K.E)}_b$

Now substitute the values we have,
$\Rightarrow \frac{1}{2}M{x}^2=\frac{1}{2}\left[\frac{1}{2}\left(\frac{M}{2}\right)y^2\right]$
Now simplify this we have,
$\Rightarrow x^2=\frac{1}{4}{y}^2$
Now take square root on both sides we have,
$\Rightarrow x=\sqrt{\frac{1}{4}{y}^2}=\frac{1}{2}y\text{. }$
Now when the man speeds up by $1 \mathrm{m}/\mathrm{s}$ then his kinetic energy becomes equal to the kinetic energy of the boy.
So the new speed of the man $=(x+1)\mathrm{m}/\mathrm{s}$.
Now according to condition we have,
New Kinetic enerav of the man = kinetic enerav of the boy

$\Rightarrow \frac{1}{2}M(x+1{)}^2=\frac{1}{2}\left(\frac{M}{2}\right)y^2$
Now let's simplify this:
$\Rightarrow (x+1{)}^2=\frac{1}{2}{y}^2$
Now, take the square root of each of our sides.
$\Rightarrow x+1=\sqrt{\frac{1}{2}{y}^2}=\frac{1}{\sqrt{2}}y$

Now from equation (1) we have, $y=2 x$ so substitute this value in equation (2) we have,
$\Rightarrow x+1=\frac{1}{\sqrt{2}}2x$
$\Rightarrow x+1=\sqrt{2}x,\left[\because \frac{2}{\sqrt{2}}=\sqrt{2}\right]$

$\Rightarrow x=\frac{1}{\sqrt{2}-1}$
Now rationalize this by $(\sqrt{2}+1)$ we have,
$\Rightarrow x=\frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$
Therefore, the needed initial speed of the guy is this $(a-b)(a+b)=\left(a^2-b^2\right)$ so we have,
$\Rightarrow x=\frac{\sqrt{2}+1}{(\sqrt{2}{)}^2-1^2}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$
So, the needed man's initial speed is this.

Hence, the correct answer is option D.