# Question #f3f56

Aug 1, 2016

$v = 2 \pm 2 \sqrt{2} \setminus \frac{m}{s}$, where $v$ is the velocity before the 2m/s increase is applied

#### Explanation:

i think you're saying than a 2m/s increase in velocity doubles KE?

if that is correct, then in math terms:

$\frac{\frac{1}{2} m {\left(v + 2\right)}^{2}}{\frac{1}{2} m {v}^{2}} = 2$, where $v$ is the velocity before the increase is applied

So cancelling some stuff

$\frac{{\left(v + 2\right)}^{2}}{{v}^{2}} = 2$

${v}^{2} + 4 v + 4 = 2 {v}^{2}$

${v}^{2} - 4 v - 4 = 0$

completing the square

${\left(v - 2\right)}^{2} - 4 - 4 = 0$

${\left(v - 2\right)}^{2} = 8$

$v = 2 \pm 2 \sqrt{2} \setminus \frac{m}{s}$

both of these solutions make sense.

if the particle is travelling to the right at $v = 2 + 2 \sqrt{2} \setminus \frac{m}{s}$ and increase its velocity to $v = 4 + 2 \sqrt{2} \setminus \frac{m}{s}$, the KE should double

Similarly, if the particle is travelling to the left, ie at $v = 2 - 2 \sqrt{2} \setminus \frac{m}{s}$, and it increases its velocity to $v = 4 - 2 \sqrt{2} \setminus \frac{m}{s}$, now moving to the right, the KE should also double.