Velocity, acceleration, displacement

[rphung]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

I honestly don't know where to begin. I tried setting up multiple equations but i keep getting more variables then equations.

 

[LowlyPion]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

I honestly don't know where to begin. I tried setting up multiple equations but i keep getting more variables then equations.

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)

 

[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

 

[LowlyPion]

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

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.