# Car acceleration/deceleration

1. Jan 29, 2006

### dxlogan187

A cab driver picks up a customer and delivers her 2.40 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration two times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.

The problems is asking for:
km traveled during acceleration
km traveled during deceleration

Since it says decceleration is two times the magnitude of the acceleration I tried multiplying 2/3 times 2.4 km to give me the correct ration of distance traveled, but that didn't work. I'm not sure where to go from here because it seems like I don't have enough information here to solve it.

Thanks,
Logan

Last edited: Jan 29, 2006
2. Jan 29, 2006

### Gamma

You have enough information to solve this. try again.

Use one of the kinematic equation for both phases of the travel.

3. Feb 5, 2010

### Late4sk00l

Is it that the magnitude is considered an X1 and X2? - Magnitude is the extent of the acceleration, what the acceleration reaches right?

4. Feb 5, 2010

### ideasrule

What's the equation for the distance traveled during constant acceleration? Get two of those formulas (one for acceleration, one for deceleration) and add them together; that will equal to the total distance driven. You also need one more equation. Do you know what it is?

5. Feb 6, 2010

### Late4sk00l

A cab driver picks up a customer and delivers her 2.00 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration is three times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases of the trip.

|--------------------------|------------------------|
ACCELERATING--------------------DECELERATING
X_0 = 0
T_0 = 0
V_0 = 0
X_1 =
T_1 =
V_1 = Speed limit
X_2 = 2000m
T_2 =
V_2 =
Acceleration can be symbolized as A
and Deceleration can be symbolized as -3A

Take note of the deceleration; if you're slowing down, then the acceleration will be a negative acceleration if you're going a positive direction. If you're going a negative direction, then the deceleration will be positive. This is important for the calculations that you will do for this problem and if you don't understand this, then you can be stuck on this problem for a long time.

Equations used

X_1 + (X_2 - X_1) = 2000 m

V^2 = V_0^2 + 2a(ΔX)

(You'll be using this equation to solve for V_1 and then use the process known as elimination.
This equation is the same thing as

-V_0^2 = -V^2 + 2a(ΔX)

-----------------------
Acceleration:

V^2 = 2a(X_1)

Deceleration:

-V^2 = 2(-3a)(2000 - X^1)

Now you can either divide the equations (Faster) or use elimination. They'll both give you the same answer. I chose to divide, you can try elimination and then check both answers.

-1 = -1/3(X_1) / (2000 - X^1)

-(2000 - X_1) = -1/3(X_1)

-2000 + X_1 = -1/3(X_1)
- X_1 -X_1

-2000 = -4/3(X_1)

-2000(-3/4) = X_1

1500 = X_1

------------------------------

X_1 + ΔX_2 = 2000m

1500 + (2000 - 1500) = 2000m

1500 + 500 = 2000

500 = X_2 which is the distance of the deceleration

1500 = X_1 which is the distance of the acceleration

I hope I helped anyone who is stuck on this problem