- #1

jgens

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## Homework Statement

A sportscar can accelerate uniformly to 120 mph in 30s. Its maximum breaking rate cannot exceed 0.7g. What is the minimum time required to go 0.5 mi assuming it begins and ends at rest?

## Homework Equations

N/A

## The Attempt at a Solution

Well, first I argued that the minimum time required to travel 0.5 mi is given by the curve v(t) where the car first accelerates uniformly at its maximum rate until some time t

_{1}and then breaks at its maximum rate until it comes to rest at some time t

_{2}. If there were some other velocity v. time curve which allowed the car to travel 0.5 mi in a smaller period of time, say u(t), then for some t' in (0,t

_{2}) we must have u(t') > v(t'). Clearly u(t) ≤ v(t) for 0 ≤ t ≤ t

_{1}since the car accelerates at its maximum rate during this period of time. Thus, v(t') < u(t') where t

_{1}< t' < t

_{2}. But since the car breaks at its maximum rate during this period for the curve v(t) and v(t

_{2}) = 0, this means that u(t

_{2}) > 0. Therefore, no such curve u(t) exists.

Now, I just need to calculate the time. To do this, I set up the equations ...

[tex]x(t_1) = \frac{a_1}{2}t_1^2[/tex]

[tex]x(t_2)-x(t_1) = (a_1t_1)(t_2-t_1) + \frac{a_2}{2}(t_2-t_1)^2[/tex]

[tex]a_1t_1 + a_2(t_2-t_1) = 0[/tex]

Where [itex]a_1[/itex] is the initial acceleration and [itex]a_2[/itex] is the breaking rate. Combining the first two equations, and using the substitution [itex]t_2 = t_1(1-a_1a_2^{-1})[/itex], we get ...

[tex]x(t_2) = t_1^2\left[\frac{-a_1}{2} + a_1\left(1-\frac{a_1}{a_2}\right) + \frac{a_2}{2}\left(1-\frac{a_1}{a_2}\right)^2 - a_2\left(1-\frac{a_1}{a_2}\right) + \frac{a_2}{2}\right][/tex]

Since [itex]x(t_2) = 1/2[/itex], we can solve this for [itex]t_1[/itex] to get [itex]t_1 = 27 s[/itex] and then use [itex]t_1[/itex] to solve for [itex]t_2[/itex] to get [itex]t_2 = 34 s[/itex].

Does this work out, or have I made a mistake somewhere along the line?