# Finding the shortest possiblee travel time

1. Sep 7, 2008

### dotfortun3

1. The problem statement, all variables and given/known data
A motorcycle can attain a maximum speed of 55 m/s, accelerate at 5 m/s^2 and decelerate at 7 m/s^2. If the motorcycle starts from a rest, find the minimum possible time it takes the motorcycle to stop exactly 800 m further down the road.

2. Relevant equations
d=rt is the only one I could think of

3. The attempt at a solution
Well, I know that I need to find the best speed to accelerate to, to be able to stop quicker and have the shortest time possible, but I really need help getting started with this

2. Sep 7, 2008

### edziura

The minimum time for the trip is such that the mc accelerates to top speed as quickly as possible, and stays there until such time as the trip is completed by decelerating to 0 speed at the maximum possible deceleration rate. The time interval and the then the distances for the beginning and ending portions of the trip can then be calculated. With this information, the time interval for the middle constant speed portion can be found.

3. Sep 7, 2008

### dotfortun3

Thank you so much! I really was just over thinking the problem and thought that I was doing it wrong! I feel dumb now! Thanks again!

4. Sep 11, 2008

### diablo2121

I have a similar question to this problem. My car can accelerate uniformly (calculated to be 1.788 m/s^2) to 120 mi/h (53.6448 m/s) in 30 s, and its maximum deceleration is -6.86 m/s^2. I'm supposed to find the minimum time to travel 0.5 mi (804.672 m) with the car beginning and ending at rest.

My problem is that the car does not reach it's maximum speed before braking since the 30 s it takes to accelerate to maximum speed travels 0.5 mi to begin with, and the car must be at rest at the end. I'm having trouble trouble setting this problem up.

5. Sep 11, 2008

### Topher925

diablo, since your maximum speed will not be reached the problem becomes a little bit more complex. I would approach the problem by creating two distance equations for the car. One for the car speeding up and one for the car slowing down. You will have two unkowns in each equation which you will then solve for.

6. Sep 11, 2008

### diablo2121

Thank you for the lead. I don't know if I got it right, but I started with two vf^2 = vi^2 + 2ax equations, one for speeding up and one for slowing down. Although I did not know vf for speeding up or vi for slowing down, I assumed the quickest would be if the car began to decelerate at a certain speed, hence my assumption is that vf and vi would be equal, a vmax if you will. Combining this with the knowledge that the distance of speeding up and the distance of slowing down ultimately added up to 0.5 mi, I set both equations to 0.5 mi and solved for a common max speed. Using this speed, I then plugged it into x = vi*t + 2at^2 to find the respective times for speeding up and slowing down from the presumed vmax.

I got an answer out of this method, but I'm not so sure that my assumptions give the minimum time it would take to drive 0.5 mi. See any flaws in my logic?