Sportscar acceleration problem

1. Nov 28, 2007

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?

2. Nov 28, 2007

Staff: Mentor

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.

3. Nov 28, 2007

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 $$t = t_1 + t_2$$?