Question

If a man increases his speed by 2m/s, his K.E. is doubled. What is the original speed of man?

Answer 1

solve further by using Quadratic formula, and u will have the required answer..
hope it helps you.

Answer 2

$\huge {\green{\underline{Solution}}}$

Let we consider it's original speed = x m/sec

and second speed be y m/sec

y = x+2 m/sec

$Kinetic \: Energy = \frac{1}{2}m v^{2}$

Takes equation first

$\red{KE_1 = \frac{1}{2}mx^{2}}$...eq(1)

Takes equation second

$\red{2KE_2 = \frac{1}{2} m(x+2)^{2}}$...eq(2)

$\blue{\boxed{KE_2 = 2 KE_1}}$

Divide both equation

$\orange{\boxed{ \frac{KE_1}{2KE_1} = \frac{ \frac{1}{2} mx^{2}}{\frac{1}{2} m(x^{2}+4+4x)}}}\\\\\frac{1}{2} = \frac{x^{2}}{x^{2}+4+4x}\\ 2x^{2} = x^{2}+4+4x\\2x^{2}-x^{2}-4-4x = 0\\x^{2}-4-4x = 0$

$x= \frac{-b+\sqrt{b^{2}-4ac}}{2a}$

or

$x=\frac{-b -\sqrt{b^{2}-4ac}}{2a}$

$x=\frac{-(-4) +\sqrt{4-4\times1\times -4}}{2\times 1}\\x=\frac{4+\sqrt{16+16}}{2}\\x=\frac{4 + \sqrt32 }{2}\\x = \frac{ \cancel 2(2 + 2\sqrt{2})}{\cancel{2}}\\x = 2+ 2\sqrt2$

$x=\frac{-(-4) - \sqrt{(-4)^{2}-4\times1\times -4}}{2\times 1}\\x=\frac{4-\sqrt{16+16}}{2}\\x=\frac{4 -\sqrt32 }{2}\\x = \frac{\cancel 2(2 +2\sqrt{2})}{\cancel{2}}\\x = 2-2\sqrt2$

$\red{\boxed{x = 2+2\sqrt2 \:or \:2-2\sqrt2}}$