Rate of change of distance between two cars

Differentiating $x^2 + y^2 = z^2$ with respect to $t$ gives $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$, which simplifies to $x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$. Here $\frac{dx}{dt} = 45$ mph and $\frac{dy}{dt} = 60$ mph are constant since the cars travel at constant speeds.

At $t=2$ hours, $x = 45(2) = 90$ miles and $y = 60(2) = 120$ miles, so $z = \sqrt{90^2+120^2} = \sqrt{8100+14400} = \sqrt{22500} = 150$ miles. Substituting into $x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$ gives $90(45) + 120(60) = 150\frac{dz}{dt}$, so $4050 + 7200 = 11250 = 150\frac{dz}{dt}$, giving $\frac{dz}{dt} = 75$ mph.

No, the separation rate changes over time and decreases as time passes. Although $\frac{dx}{dt} = 45$ and $\frac{dy}{dt} = 60$ are constants, the rate $\frac{dz}{dt}$ is determined by $\frac{dz}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{z} = \frac{45x + 60y}{z}$. Since $x = 45t$ and $y = 60t$ grow linearly with $t$, while $z = \sqrt{x^2 + y^2} = 75t$ also grows linearly (maintaining the 3-4-5 ratio), we can write $\frac{dz}{dt} = \frac{45(45t) + 60(60t)}{75t} = \frac{2025t + 3600t}{75t} = \frac{5625t}{75t} = 75$ mph for all $t > 0$. Therefore, the separation rate is actually constant at 75 mph.