Minimum time required to cover distance, beginning and ending at rest

[Fantini]

This is problem 1.16 from Kleppner & Kolenkow's Introduction to Mechanics. I didn't know whether to post it here or pre-university math.

A sportscar, Fiasco I, can accelerate uniformly to $120$ mi/h in $30$ s. Its maximum braking rate cannot exceed $0.7g$. WHat is the minimum time required to go $\frac{1}{2}$ mi, assuming it begins and ends at rest? (Hint: A graph of velocity vs. time can be helpful.)

My thought process: Since it can accelerate uniformly to 120 mi/h in 30 seconds I've converted this to an acceleration of
$$\frac{120 \frac{\text{mi}}{3600 \text{ s}}}{30 \text{ s}} = \frac{1}{30} \frac{1}{30} \frac{ \text{mi}}{\text{s}^2} = \frac{1}{900} \frac{\text{mi}}{\text{s}^2}.$$
Since the maximum braking rate cannot exceed $0.7g$ then its maximum deacceleration cannot exceed $0.00609 \text{ mi}/\text{s}^2$. Assuming that it can deaccelerate at the maximum rate instantaneously, graphing the velocity vs. time as in the hint we'd have something like this:[desmos="-1,2,-1,2"]y=x;y=5-3x[/desmos]
Desmos's merely illustrative. The distance covered is the area of the triangle. Denoting the velocity it may achieve as $v_0$, the time taken to reach velocity $v_0$ as $t_0$ and the time taken for everything to occur as $t_1$, we can write a few relations. We know
$$\frac{v_0}{t_0} = \frac{1}{900} \frac{\text{mi}}{\text{s}^2},$$
$$\frac{v_0}{t_0-t_1} = - 0.00609 \frac{\text{mi}}{\text{s}^2},$$
$$\frac{v_0 t_1}{2} = \frac{1}{2} \text{ mi}.$$

Writing
$$v_0 = \frac{t_0}{900} \text{ and } v_0 = 0.7g (t_1-t_0) = \frac{t_0}{900}$$
we can obtain $t_0$ in terms of $t_1$ as
$$t_0 = \frac{0.7g t_1}{\frac{1}{900} + 0.7g}$$
and $v_0$ in terms of $t_1$ as
$$v_0 = \frac{0.7g t_1}{1 + 900 \cdot 0.7g}.$$
Combining these in the last equation we get
$$\frac{0.7g t_1^2}{1 + 0.7g} = 1 \text{ and } t_1 = \sqrt{ \frac{1}{0.7g}+ 900}.$$
This is approximately $t_1 \approx 33,7$ seconds. It feels right because the distance covered by the sportscar if it maintained uniform acceleration for $30$ seconds is exactly $\frac{1}{2}$ mile.

Is this correct?

 

[mfb]

This is a sloppy use of units and 0.006 mi/s² is 1 g, not the maximal 0.7 g, but apart from that it is right.

Cross check: From 120 mi/h the car would need 7.8 seconds to stop, so we expect the car to accelerate for ~26 seconds. The last ~8 seconds are traveled at half the top speed (on average) to cross the distance the car would travel in 4 seconds with an ongoing acceleration.