Question

If on increasing the speed of a vehicle by 2$m{s}^{-1}$ , the kinetic energy gets doubled ,then what would be its initial speed?

Answer 1

Let the initial speed of the vehicle be $u$

and given final velocity of the vehicle $v=u+2$

Now initial kinetic energy $E_{Ki}=\frac{1}{2}m{v}^2$

and as per question final kinetic energy$E_{Kf}=2E_{Ki}$

Thus we can write final kinetic energy of the vehicle as,

$2E_{Ki}=\frac{1}{2}m(u+2{)}^2$

$2\left(\frac{1}{2}\right)$ $\left(m{u}^2\right)$=$\frac{1}{2}$ $m{u}^2$ +4u+4

$2u^2-u^2-4u-4=0$

$u^2-4u-4=0$

Now applying sri ardhachariya rule we get

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

$u=\frac{-(-4)\sqrt{(-4{)}^2-4(-4)c}}{2}$

$=\frac{4+5.56}{2}$

$=4.825$