Constant power, variable speed

From the definition of Power in Mechanics, we can write the relation that:- $P = Fv$ $P = mav$ Using the relation that $a = v\dfrac{dv}{dx}$ . $P = mv\dfrac{dv}{dx}v$ $Pdx = mv^{2}\dfrac{dv}{dx}$ Integrating the above expression we get:- $\int_{0}^{x} Pdx = \int_{0}^{v} mv^{2}dv$ $Px = \dfrac{mv^{3}}{3}$ $\dfrac{3Px}{m} = v^{3}$ $\dfrac{dx}{dt} = \Big( \dfrac{3Px}{m}\Big) ^{\frac{1}{3}}$ $\dfrac{dx}{x^{\frac{1}{3}}} = \Big( \dfrac{3P}{m} \Big)^{\frac{1}{3}} dt$ Again integrating both sides we get:- $\int_{0}^{x}\dfrac{dx}{x^{\frac{1}{3}}} = \int_{0}^{t} \Big( \dfrac{3P}{m}\Big)^{\frac{1}{3}} dt$ $\dfrac{3x^{\frac{2}{3}}}{2} = \Big( \dfrac{3P}{m}\Big)^{\frac{1}{3}} t$ Now cubing both sides:- $\dfrac{27x^{2}}{8} = \dfrac{3Pmt^{3}}{m}$ $x^{2} = \dfrac{8Pt^{3}}{9m}$ $x = \dfrac{2\sqrt{2}}{3} \sqrt{\dfrac{P}{m}}t\sqrt{t}$ Now substituting the given values we get $\boxed{ x = 180m}$

Assuming that the power is entirely used for kinetic energy, we have $K(t) = W(t) = P\cdot t= {10}^{5}t$ where $K$ is the kinetic energy of the car in Joules. From $K = \frac{1}{2}mv^2$ , we have $10^{5}t = \frac{1}{2}\cdot 2000v^2 \implies |v| = 10\sqrt{t}$ . To find the distance travelled from $t = 0$ to $t = 9$ , we integrate the speed: $\Delta x = \displaystyle\int\limits_{t=0}^{t=9} {|v|} dt = \displaystyle\int\limits_{t=0}^{t=9} {10\sqrt{t}dt} = 10\times\frac{3}{2}(9^\frac{3}{2} - 0^\frac{3}{2}) = 180$ m.

Remember the definition of power, and that the only energy that is being added to our system is kinetic:

$P \ = \ \frac{dE}{dt} \ = \ \frac{d}{dt} \ \frac{1}{2} \ mv^2 \ = \ \frac{1}{2} \ m \ 2v \ \frac{dv}{dt} \ = \ m v \frac{dv}{dt}$

We then rearrange, and solve the differential equation for velocity (remember that power is constant):

$\frac{P}{m} \ = \ v \frac{dv}{dt} \ \Rightarrow \ \displaystyle\int \ \frac{P}{m} \ dt \ = \ \displaystyle\int \ v \ dv \ \Rightarrow \ \frac{v^2}{2} \ = \ \frac{Pt}{m} \ \Rightarrow \ v(t) \ = \ \sqrt{\frac{2Pt}{m}}$

Finally, we integrate with respect to time from $0$ to $9$ :

$d \ = \ \displaystyle\int_{0}^{9} \ v(t) \ dt \ = \ \displaystyle\int_{0}^{9} \ \sqrt{\frac{2Pt}{m}} \ dt \ = \ \sqrt{\frac{2P}{m}} \ \frac{2t^{3/2}}{3} \ \biggr\rvert_{0}^{9} \ = \ 180 \ m$

$f\to \frac{p}{v}$ , $v\to x'(t)$ , $x''(t)=\frac{p}{m\ x'(t)},x'(0)=0,x(0)=0 \Rightarrow x(t)\to \frac{2 \sqrt{2} (p\ t)^{3/2}}{3 \sqrt{m}\ p}$ , substituting p as 100 kw, m as 2000 kg and t as 9 s and converting to meters: 180 meters.

good one ....if original