How to Calculate Minimum Time for a Sportscar to Travel Half a Mile?

[tronter]

A sportscar can accelerate uniformly to 120 \frac{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 that it begins and ends at rest.

So 120 = 0 + 30a, \ a = 4.

Then \frac{1}{2} = 2t^{2}. But I know I have to incorporate the braking rate.

How would I proceed from here?

 

[Astronuc]

Make sure that the accelerations are in the same units.

120 mph = 176 ft/s and the acceleration from 0 to 176 ft/s in 30 s is 5.867 ft/s², and the magnitude of deceleration is 0.7*9.81 ft/s² = 6.867 ft/s².

Now over 0.5 mile or 2640 ft, if the car accelerates over distance d ft, then it must decelerate over distance (2640 - d) ft, and one must find d such that t is minimized, or the average speed is maximized since v(avg) = 2640 ft/t, where t is the time to travel 1/2 mile.

 

[tronter]

So then we have at_1 - 0.7gt_2 = 0 and 2640 = \frac{1}{2}(at_{1}^{2} - 0.7gt_{2}^{2}) and solve for t_1 and t_2?So then t = t_1 + t_2?