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Two cars passing each other (constant velocity)

  • Thread starter praecox
  • Start date
17
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1. Homework Statement

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 220 m. If the red car has a constant velocity of +20 km/h, the cars pass each other at x = 44.6 m, and if it has a constant velocity of +40 km/h, they pass each other at x = 76.9 m.

(a) What is the initial velocity of the green car? (Indicate direction with the sign of your answer.) in m/s

(b) What is the acceleration of the green car? (Indicate direction with the sign of your answer.) in m/s2

2. Homework Equations

I think I'm supposed to use:
Xf = Xi + Vxi*t + 0.5at2

3. The Attempt at a Solution

I'm having a hard time with this one - even knowing where to start.
For part (a), it suggested setting the equations equal to each other for each set of data. I converted the constant velocity to m/s (20 km/hr = 5.56 m/s; 40 km/hr = 11.11 m/s) to avoid conversion errors later.

So I figured out how much time it would take the car at 5.56 m/s to reach the 44 m mark:
∆x/V = t --> 44.6m / 5.56m/s = 8.02 sec. So I thought I could take the green car from the opposite end and see how fast it would have to go to clear the distance and meet the red car at 8.02 seconds. So ∆x=-175.4m (accounting for direction), t=8.02 and V is unknown.

When I plug that into the equation I get ∆x/t = V (-175.4/8.02), so V=-21.87m/s... but when I submitted the answer online it was wrong and it said the right answer was -13.3m/s.

And then for part (b), I thought if you have a constant velocity, you have an acceleration of 0. As far as the problem seemed to me, both cars are moving at a constant velocity when the problem "starts". But my homework program says the answer is -2.12 m/s^2. so... huh?

Any help would be awesome, guys! :biggrin:
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
319
0
well, here is my solution
the red car has a constant velocity. let's suppose that it moves along the x-axis in positive direction, so it's position function is something like r(t) = vt + a. since it is at 0 at the time t=0 then the coefficent a is zero. the green car has a constant acceleration, so its position is given by a function like g(t) = [tex] 1/2at^2 + wt + b[/tex]. since the green car is at 220 m at time t=0 the coefficient b = 220. now if we substract the two equations we will have the equation below:

[tex] \Delta x = 1/2 at^2 + (w-v)t + 220. [/tex]

now we can write down the following equations with given data:
44.6 = 5.56 t -> t ~ 8 seconds
76.9 = 11.12t -? t ~ 6.9 seconds

at the moment when the two cars are passing each other, the distance between them ([tex] \Delta x [/tex]) equals zero. so we will have the following equations:
0 = 32a + (w-5.56)8 + 220
0 = 23.805a + (w-11.12)6.9 + 220

now we have a simple LE system that we can solve for a and w.
-175.52 = 32a + 8w
-143.272 = 23.805a + 6.9w

a ~ -8114/3795 ~ -2.13 m/s2
w ~ 508063/37950 ~ -13.38 m/s

the minus signs mean that the green car is moving in negative direction of the x axis (because we supposed that the red car is moving along positive direction of the x axis).

note that the time variable is the same for both cars. that means when the two cars are passing each other they are at the same t.
 
Last edited:
17
0
That's a great explanation... but I'm still a little confused.
I see how you got the final equation that we're going to use, but what do you mean "LE system"? I'm not sure how to solve for two variables at the same time... :/
 
319
0
That's a great explanation... but I'm still a little confused.
I see how you got the final equation that we're going to use, but what do you mean "LE system"? I'm not sure how to solve for two variables at the same time... :/
by a simple LE system I meant a system of linear equations of two unknowns. you can solve them using elimination or you can think of them as two lines intersecting at some point, that point is the answer we're looking for.
read this article on wikipedia:
http://en.wikipedia.org/wiki/System_of_linear_equations

if you are familiar with matrices you can use this online tool to solve your systems of linear equations when the numbers are freaky:
http://www.uni-bonn.de/~manfear/solve_lineq.php [Broken]
 
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