A car is slowing down at a constant pace

[Avathacis]

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

Homework Statement

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?

Homework 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?)

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? :(

 

[tiny-tim]

Hi Avathacis!

A car is slowing down at a constant pace. Half of it's stopping path (i'm not sure what do you call it in english) the car travels in t₁=4s. In how much time since the stopping started the car stops?
…
v=v₀t_?+at_?² (i'm guessing that v=0?)
v₀=at_?
v₁=a(t₂-t₁) (?) …

("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.

 

[Avathacis]

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.

<br /> v^2 = v_1^2 + 2 a \Delta x?<br />
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)
<br /> v_1 = root - 2 a S<br /> 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 :>.

 

[Doc Al]

<br /> v^2 = v_1^2 + 2 a \Delta x?<br />
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?

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.)

 

[Avathacis]

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.)

v₀=at₁ (?)
v₁=a(t₂-t₁)

 

[Doc Al]

v₀=at₁ (?)

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

v₁=a(t₂-t₁)

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.

 

[Avathacis]

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.

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 :<.

 

[Doc Al]

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

Sure. That time is what you call T₂--the total time.

Though I'm not sure i get what you mean by magnitude

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:

V_f = V₀ - at

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

 

[Avathacis]

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:

V_f = V₀ - at

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

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

 

[Doc Al]

Kinda like modules? (|-11|=11?)

Exactly like that.

I guess you're kinda turning to the point that I'm making a + or - mistake somewhere right?

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.

 

[Avathacis]

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.

<br /> S_1 = x_0 + v_0t_2 + at_2^2/2<br />

<br /> S_2 = x_0 + v_1(t_2 - t_1) + a(t_2 - t_1)^2/2<br />

<br /> v_0 = at_2<br />

<br /> v_1 = a(t_2 - t_1)<br />

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

 

[Doc Al]

<br /> S_1 = x_0 + v_0t_2 + at_2^2/2<br />

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

<br /> S_2 = x_0 + v_1(t_2 - t_1) + a(t_2 - t_1)^2/2<br />

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

<br /> v_0 = at_2<br />

<br /> v_1 = a(t_2 - t_1)<br />

OK.

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

Yes, then set S1 = S2.

 

[Avathacis]

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.

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

 

[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?

 

[Doc Al]

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

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.