We'll show that $$\frac{x}{{10}^7}=e^{-\left(\frac{y}{{10}^7}\right)},$$ The point $Q$ moves at a constant speed of ${10}^7$, so we have $dy/dt={10}^7.$ Since $P$ moves at a speed that is proportional to the distance $x$ left to travel we have $dx/dt=-x.$ From this we see that $$dy/dx=-\frac{{10}^7}{x},$$ which gives $$y=-{10}^7\ln cx,$$ for some constant $c.$ We can work out the value of $c$ using our initial conditions. At the start, the point $P$ still needs to travel the whole length of the line segment $AB$, which is ${10}^7$. Therefore $x_0={10}^7$. The point $Q$ hasn't gone anywhere yet, so $y_0=0.$ Plugging this into the expression above gives $$0=-{10}^7\ln c{10}^7,$$ so $$e^0=1=c{10}^7.$$ Therefore, $$c=\frac{1}{{10}^7},$$ so $$y=-{10}^7\ln \frac{x}{{10}^7},$$ and $$\frac{x}{{10}^7}=e^{-\left(\frac{y}{{10}^7}\right)},$$ which is what we wanted to show.

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