Question

If a man increases his speed by $2\,m{s^{ - 1}}$, his K.E. is doubled. The original speed of the man is:

Answer 1

Hint:Here we will use the formula of kinetic energy to calculate the original speed of the man. Here, the kinetic energy after attaining the final speed will be double the kinetic energy at the initial speed. After solving the ratio of kinetic energies we will get a polynomial equation and we will use the quadratic formula to solve this polynomial.

Formula used:
The formula of the kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
Here, $K$ is the kinetic energy, $m$ is the mass, and $v$ is the velocity.
Also, the quadratic formula used for solving the polynomial is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a$ is the constant of ${x^2}$ , $b$ is the constant of $x$, and $c$ is the constant.

Complete step by step answer:
Let $u$ is the initial velocity of a man. Therefore, the kinetic energy of the man at the initial velocity $u$ is given by
${K_1} = \dfrac{1}{2}m{u^2}$
Now, when the man increases his speed by $2\,m{s^{ - 1}}$ , therefore, his final speed or velocity will become $v = u + 2$ . Also, at this stage, the kinetic energy will become double. Hence, the kinetic energy due to the final velocity of the particle is given by
$2{K_1} = \dfrac{1}{2}m{\left( {u + 2} \right)^2}$
Now, dividing ${K_1}$ by ${K_2}$ , we get
$\dfrac{{{K_1}}}{{2{K_1}}} = \dfrac{{\dfrac{1}{2}m{u^2}}}{{\dfrac{1}{2}m{{\left( {u + 2} \right)}^2}}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{{{u^2}}}{{{{\left( {u + 2} \right)}^2}}}$
$ \Rightarrow \,{\left( {u + 2} \right)^2} = 2{u^2}$
We will solve the above equation in the form of a polynomial.
${u^2} + 4 + 4u = 2{u^2}$
$ \Rightarrow \,{u^2} - 4u - 4 = 0$
Hence, the above polynomial is the required polynomial. Now, to solve this equation, we will use the quadratic formula which is given by
$u = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, putting the values, we get
$u = \dfrac{{ - ( - 4) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 1 \times \left( { - 4} \right)} }}{{2 \times 1}}$
$ \Rightarrow \,u = \dfrac{{4 \pm \sqrt {16 + 16} }}{2}$
$ \Rightarrow \,u = \dfrac{{4 \pm \sqrt {32} }}{2}$
$ \Rightarrow \,u = \dfrac{{4 \pm 4\sqrt 2 }}{2}$
$ \Rightarrow \,u = 2 \pm 2\sqrt 2 $
$ \therefore \,u = 2 + 2\sqrt 2 $ or $u = 2 - 2\sqrt 2 $
Here, we will neglect $u = 2 - 2\sqrt 2 $ because the speed can never be less than zero.

Therefore, the original speed of the man is $2 + 2\sqrt 2 $.

Note:Remember, we have taken the final velocity as $v = u + 2$ because it is given in the question that the speed of the man is increased but not the final velocity is given in the question. Also, the kinetic energy will become double the previous kinetic energy. The original speed of the man will be the initial speed.