Velocity vs. Time with Uniform Acceleration and Braking Rate

[leejqs]

Homework Statement

A sportscar, Fiasco I, can accelerate uniformly to 68 m/s in 30 s. Its maximum braking rate cannot exceed 0.75g. What is the minimum time required to go 1300 m, assuming the car begins and ends at rest?
(Hint: A graph of velocity vs. time can be helpful.)

Homework Equations

No "equations" really, rather, just the calculus relationship between position, velocity, and acceleration.

The Attempt at a Solution

In constructing a velocity vs. time graph, I found the acceleration (or slope of the line) to be 68m/s / 30 s = 2.27m/s². Likewise, the acceleration (or slope) when the car is decelerating is .75*9.81m/s²=-7.36m/s^s.

Obviously, the area under the curve of this graph is the distance traveled by the car. I calculated that in order for the car to travel 1300m at its constant acceleration of 2.27m/s², it would need to travel for 33.9 seconds. [The integral from 0 to 33.9 of 2.27t = 33.9].

This above calculation is probably irrelevant to the problem at hand, but I have NO CLUE as to where the car should stop accelerating and put on its brakes. Is this a differential equation problem possibly? Somehow I need to figure out the maximum time the car can accelerate in order for it to slam on its brakes and skip to a stop so that its total distance traveled is exactly 1300m. What calculus is used to find this mid-point of acceleration to deceleration?

Any help would be greatly appreciated! Thanks!

 

[rl.bhat]

In the problem duration of acceleration t1 is given. It moves with uniform velocity, which is given, during the time t2. It decelerates during time t3, which you can find.
So the distance traveled d = The area of the trapezium = 1/2*( t1 + 2*t2 + t3)* velocity.

 

[jbunniii]

I was thinking about this same problem recently. It is from Kleppner and Kolenkow's "An Introduction to Mechanics," problem 1.16 and in that book, it reads as follows:

A sports car, the Fiasco I, can accelerate uniformly to 120 mi/h in 30 sec. Its maximum braking rate cannot exceed 0.7g. What is the minimum time required to go 1/2 mile, assuming it begins and ends at rest? (Hint: A graph of velocity vs. time can be helpful.)

The hint is clear enough: the velocity versus time curve, v(t), must satisfy the following constraints:

1) v(0) = 0
2) v(T) = 0 where T > 0 is the end time which you want to make as small as possible
3) the area under the curve = total distance traveled, must equal the given distance D.
4) The derivative function, which is the acceleration a(t), must satisfy B <= a(t) <= A for all t, where B = the maximum deceleration and A = the maximum acceleration.
5) Implicitly, v(t) must be a continuous function of time.

The task is then to find the function v(t) that minimizes T, subject to all of these constraints.

It seems intuitively plausible that the optimal function v(t) is a triangle, with upward and downward slopes equal to the maximum acceleration and braking deceleration, respectively. In that case, the only parameter available for varying is T1, the time at which the drive instantly switches from maximum acceleration to maximum braking. This parameter in turn can be solved for given that the triangle must have a specific area.

The subtler point that I haven't been able to prove to myself is that the "triangle" strategy is indeed the optimal one. i.e., How can I establish that constant acceleration followed immediately by constant deceleration is the best solution, rather than some more complicated curve? Again, it seems intuitively right, but to prove it mathematically seems to fall under the domain of the calculus of variations. But this is not very familiar ground for me, and I'm a bit surprised to see it show up in a problem in the first chapter of an introductory physics book.

Without much difficulty, I was able to show that if the figure is constrained to be a triangle of specific area, and with upward slope 0 <= a <= A, and downward slope 0 >= b >= B, then the optimal time is achieved by setting the rising slope to be the maximum acceleration A, and the falling slope to be the maximum braking deceleration B, but this is pretty obvious anyway. What's not so obvious is that the optimal figure is a triangle.

 

[rl.bhat]

In the first part, initial velocity, final velocity and time is given. From these data you can find the acceleration and displacement. In the second part, initial velocity is the same as the final velocity in the first part. Maximum acceleration is given. Final velocity is zero. Find the dispalcement. Add the displcements. If this displacement is equal to 1/2 mile, then the graph is triangular. Otherwise it is trapezium.

 

[jbunniii]

There's only one part to the problem:

A sports car, the Fiasco I, can accelerate uniformly to 120 mi/h in 30 sec. Its maximum braking rate cannot exceed 0.7g. What is the minimum time required to go 1/2 mile, assuming it begins and ends at rest? (Hint: A graph of velocity vs. time can be helpful.)

Initial velocity, final velocity, and distance traveled are given. Maximum acceleration and deceleration are given. Subject to these constraints, minimize the time from start to finish.

I don't believe that a trapezoid would ever be the optimal solution.

My reasoning: suppose you have a trapezoidal profile starting at time 0 and ending at time T. Now take a small section of the flat top of the trapezoid. Replace it with a small triangular peak that rises above the flat top. You have therefore added some area. Now delete the same amount of area by shortening the remainder of the flat section. Thus the end time T decreases, so you have just found a better solution.

 

[rl.bhat]

In the problem the maximum decelaration is given, not the maximum accceretion.

 

[jbunniii]

In the problem the maximum decelaration is given, not the maximum accceretion.

Both are given:

A sports car, the Fiasco I, can accelerate uniformly to 120 mi/h in 30 sec. Its maximum braking rate cannot exceed 0.7g.

 

[rl.bhat]

In the problem, the acceleration is constant. But the braking deceleration is variable. And its maximum value is 0.75*g.