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

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The discussion revolves around calculating the minimum time required for a sportscar, Fiasco I, to cover a distance of half a mile while starting and ending at rest. The car can accelerate uniformly to 120 mi/h in 30 seconds and has a maximum braking rate of 0.7g. The calculations involve determining the acceleration and deceleration rates, leading to the formulation of time relationships based on velocity and distance. The derived equations suggest that the total time required to complete the distance is approximately 33.7 seconds, which aligns with the expected performance of the car. The analysis confirms the feasibility of the calculated time, considering the car's acceleration and braking capabilities.
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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?
 
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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.
 
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