# Velocity, acceleration, displacement

1. Oct 5, 2008

### rphung

1. The problem statement, all variables and given/known data
a cab driver picks up a customer and delivers her 2.00km away, on a straight route. The driver accelerates to the speed limit and, on reaching it, begins to decelerate at once. the magnitude of the deceleration is three times the magnitude of the acceleration. find the lengths of the acceleration and deceleration phases.

2. Relevant equations
vf=vi+at
x=xi+vit+.5at^2

3. The attempt at a solution
I honestly dont know where to begin. I tried setting up multiple equations but i keep getting more variables then equations.

2. Oct 5, 2008

### LowlyPion

Consider using the equation V2 = 2*a*x
The acceleration phase is x and the deceleration distance is (2 - x) right?
Since the V is the same:

2*a*x = 2*3a*(2-x)

3. Oct 5, 2008

### rphung

can you explain why the v's are the same?

I thought the equation was vf^2=vi^2+2ad and when the car starts deaccelerating the vi is some unknown velocity while the vf will be 0

4. Oct 5, 2008

### LowlyPion

Sure.

You have an acceleration phase. It gets to the speed limit from 0. That would be the first equation.
The second is from the speed limit back to 0. Same speed limit. Same speed. Then you can set the two equal.