# Homework Help: A car is slowing down at a constant pace

1. May 23, 2010

### Avathacis

A car is slowing down at a constant rate...

1. The problem statement, all variables and given/known data
WARNING: This is a translation. If you find something you can't understand, i'll try to explain.
A car is slowing down at a constant rate. The car travels half of its stopping distance in t1=4s. How long does the car take to stop?

2. Relevant equations
t1=4s
S1=S2

S1 - first part of the path
S2 - second part of the path
v0 - starting speed
v1 - speed after the first half of the path
t2 - whole time spent traveling
x0 - starting position (0?)

3. The attempt at a solution
Err:
v=v0t?+at?2 (i'm guessing that v=0?)
v0=at?
v1=a(t2-t1) (?)

So what i am guessing is that i should make
S1=x0+v0t?+at?2/2
S2=x0+v1(t2-t1)+a(t2-t1)2/2

and then put in the v0 and v1 i got earlier. And then since S1=S2 i put what i got from the above.. I think the a's should get removed if i divide by "a" and i should get a quadratic equation. Here's the problem, i've tried doing it (filled around 3 pages of my notes) and i get the wrong answer. Which, i think, is mainly due to the fact i don't know what times to put in where. The answer should be:
t22-4t2t1+t12=0

And i get always something around that ^. But never it. Sometimes i'm missing just 1 number, sometimes a letter. What to do? :(

Last edited: May 23, 2010
2. May 23, 2010

### tiny-tim

Hi Avathacis!
("A car is slowing down at a constant rate. The car travels half of its stopping distance in t1=4s. How long does the car take to stop?" )

Hint: find v1 by using a formula which doesn't involve t.

3. May 23, 2010

### Avathacis

$$v^2 = v_1^2 + 2 a \Delta x?$$
I don't think i'm supposed to know this formula :>. Or at least i haven't used it. Or i'm not getting something?
Also thanks for the fixes.

Wouldn't that make (assuming i am correct about v being 0)
$$v_1 = root - 2 a S$$ Also i assume that delta x is S in my problem? :>

Also could anyone tell me if my guesses are correct and why it would be great :>.

Last edited: May 23, 2010
4. May 23, 2010

### Staff: Mentor

You don't need that formula. Just use your equations for distance for each half of the distance.

Hint: Express both V1 and V0 in terms of a, T1, T2. (Careful with signs.)

5. May 23, 2010

### Avathacis

v0=at1 (?)
v1=a(t2-t1)

6. May 23, 2010

### Staff: Mentor

This one is not quite right. V0 goes to 0 in time T2.
Good.

Note that here you are using 'a' to represent the magnitude of the acceleration. I suggest that you revise your earlier equations to use the same convention.

7. May 23, 2010

### Avathacis

1) I'm supposed to use the time the speed gets down to 0? It'll be helpful!

Though i'm not sure i get what you mean by magnitude :<.

8. May 23, 2010

### Staff: Mentor

Sure. That time is what you call T2--the total time.
Magnitude means the quantity without regard for its sign. For example: The magnitude of the number -11 is just 11.

So if you call the magnitude of the acceleration 'a', then the actual acceleration will be -a, since it's slowing down. For example:

Vf = V0 - at

Applying that to the final speed, we get:
0 = V0 - at2
or
V0 = at2

9. May 23, 2010

### Avathacis

Kinda like modules? (|-11|=11?)
I guess you're kinda turning to the point that i'm making a + or - mistake somewhere right?

10. May 23, 2010

### Staff: Mentor

Exactly like that.
It's hard to tell. Why don't you write down your four equations, using 'a' as the modulus of the acceleration. (Thus, you may have to put '-a' in some of your formulas.) You'll have your two distance equations and the two velocity equations.

11. May 23, 2010

### Avathacis

$$S_1 = x_0 + v_0t_2 + at_2^2/2$$

$$S_2 = x_0 + v_1(t_2 - t_1) + a(t_2 - t_1)^2/2$$

$$v_0 = at_2$$

$$v_1 = a(t_2 - t_1)$$

Right, i know i have to put in the v_0 and V_1 into S_1 and S_2.

12. May 24, 2010

### Staff: Mentor

Since S1 is the first part of the distance, shouldn't the time be t1, not t2?

Since the car is slowing down, you need to replace 'a' with '-a' in both of those equations.

OK.

Yes, then set S1 = S2.

13. May 24, 2010

### Avathacis

I think i can solve it now. Well explained, thanks. I will post again if i need more help :>.

14. May 24, 2010

### Avathacis

I have a quick question:
After solving the equation a little i get:
t2=2t2t1+t2t1sqrt2
Can i divide by t2 to remove it?

15. May 24, 2010

### Staff: Mentor

Algebraically, sure you can. But that equation cannot be correct. (You cannot have a physically meaningful equation where terms have different units. You can't have a time term equal to a time-squared term.)

So go over it again carefully.