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# Physics Problem: Driving Kinematics

snowboardalliance
Well, it's been a year since I took basic physics and I need some help. I'm working on a traffic simulation for a research job I have and I'm getting stuck trying to figure out the best way to do something.

Basically, I have two cars distance d apart each with a velocity and constant acceleration (well I'll call it constant for simplicity).

Now, if car number 1 is behind car number 2 and is traveling faster but has a negative acceleration, can it slow down before it hits car number 2?

So there is:
car1: s1 (position), v1, a1
car2: s2, v2, a2
d

conditions: s1 < s2, a1 < 0, and v1 > v2

I remember doing problems like this in physics, but I'm stuck.

EDIT:

I think I've figured it out,
the time to reach the same velocity is:
 Code: v1 - v2 t = _______     a2 - a1

because
 Code: vf = vi + at

and if I just check that the distance of car1 at time t is less than the distance of car2 at time t + initial distance, then I know that the cars will not collide.

My problem is, if the accelerations are the same, it's division by 0 and I can't figure out an alternative equation or what that would really mean.

EDIT2:

Ok, I guess my calculation for time isn't really the time I want. What time would I need to plug in to:
 Code: s = vi*t + 1/2at^2

To see if the car behind would hit the car in front?
Xanatos
Your problem is in your equation for t. Remember that your originally stated that car #1 has a negative acceleration. Therefore the accelerations are never the same. Since a1 will always be negative you never have a problem dividing by zero. You could just as easily right

nanunath
u know this eqn:
s=ut+[.5a(t^2)] ...(I)

So:
Assume:Let after time 't'[if t=0 is the instant @ which u've given values] the cars hit eachother..

Let car 1 travel 'd1' and car2 travel 'd2' dist from t=0 b4 they hit @ t=0

So condition for collosin is:
s1+d1=s2+d2 ...(II)

d1=[(v1) + (.5)(a1)(t^2)] ...from (I) and a1<0
and
d2=[(v1) + (.5)(a1)(t^2)] ...a2>0

Dun!..
Subst d1,d2 in Eqn(II)..only t is unkown...solve quadratic..if t is real...they will collide...if t is imaginary...they'll never collide...
infinisa
Hi snowboardalliance

nanunath has the right idea, but the working had a few errors.

The equations
 Quote: s1+d1=s2+d2 ...(II) d1=[(v1) + (.5)(a1)(t^2)] ...from (I) and a1<0 and d2=[(v1) + (.5)(a1)(t^2)] ...a2>0

should be:
s1+d1=s2+d2 ...(II)

d1=[(v1)t + (.5)(a1)(t^2)] ...from (I) and a1<0
and
d2=[(v2)t + (.5)(a2)(t^2)] ...a2>0

Substituting for d1 and d2 in (II) gives
s1 + (v1)t + (.5)(a1)(t^2) = s2 + (v2)t + (.5)(a2)(t^2)

So
(.5)(a2 - a1)(t^2) + (v2 - v1)t + (s2 - s1) = 0

This is a quadratic in t, so has a real solution if the discriminant "b^2 - 4ac" is >= 0
i.e. if
(v2 - v1)^2 - 2(a2 - a1)(s2 - s1) >=0

(v2 - v1)^2 >= 2(a2 - a1)(s2 - s1)

We also need to make sure t>=0, since a solution with t<0 is a reference to the past, not the future.

For this, note that the coefficient of t^2 > 0, the coefficient of t < 0 and the constant is > 0.
This tells us that the product and the of the solutions is positive, so (if real) they're both positive.

And there you have it!
nanunath
^ Right my friend...
error is in that fundamental eqn...shit yaar.. ..I missed out that "t" in "s1=u1*t + (.5*a1*t1*t1)"
neway..Ok..coz I typed that @2.00AM...
infinisa
 nanunath wrote: ^ Right my friend... error is in that fundamental eqn...shit yaar.. ..I missed out that "t" in "s1=u1*t + (.5*a1*t1*t1)" neway..Ok..coz I typed that @2.00AM...

Do I get that
I wrote my solution at about 1:45 am, but just to be on the safe side, I checked and posted it the next morning!
nanunath
^hmmmmm
.... ur smart my friend .. unfortunately ... I'm not ... ... ...
infinisa
 nanunath wrote: ^hmmmmm .... ur smart my friend .. unfortunately ... I'm not ... ... ...

"Smart" is not about not making mistakes - it's about learning from them.
My Dad used to say: "He who never makes a mistake never makes anything"
nanunath
Hhhmm.u r really smart ..my frnd!...I swear!
snowboardalliance