Two cars passing each other (constant velocity)

In summary, two identical cars, one red and one green, are moving towards each other on adjacent lanes parallel to an x axis. The red car has a constant velocity of +20 km/h and the green car has a constant acceleration. At different points along the x axis, the cars pass each other at x = 44.6 m and x = 76.9 m respectively. To find the initial velocity and acceleration of the green car, a system of linear equations is set up using the given data and solved for the unknowns. The initial velocity of the green car is -13.38 m/s and the acceleration is -2.13 m/s^2.
  • #1
praecox
17
0

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

Homework Equations



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

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:
 
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  • #2
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.
 
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  • #3
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... :/
 
  • #4
praecox said:
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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  • #5

Your approach to part (a) is correct, but you made a mistake in your calculations. When you take the green car's initial position as -175.4 m, you should also change the sign of the distance traveled by the red car to +175.4 m. This will give you a positive value for the velocity of the green car, which is the correct answer.

For part (b), you are correct that the acceleration is 0 if the cars are moving at a constant velocity. However, in this problem, the green car is accelerating because it has to change its velocity in order to pass the red car. The acceleration of the green car can be calculated using the equation a = (Vf - Vi)/t, where Vf is the final velocity (11.11 m/s) and Vi is the initial velocity that you calculated in part (a). This will give you a negative acceleration, which is the correct answer.

Remember, acceleration is the rate of change of velocity, so even if the cars are moving at a constant velocity, the green car still has to accelerate in order to pass the red car. I hope this helps clarify the problem for you.
 

What is the concept of "two cars passing each other" with constant velocity?

Two cars passing each other with constant velocity refers to a situation where two cars are traveling in opposite directions on a road at a constant speed.

What is the importance of studying two cars passing each other with constant velocity?

Studying two cars passing each other with constant velocity can help us understand the principles of motion, such as velocity, acceleration, and distance. It can also provide insights into traffic flow and safety on roads.

How do you calculate the velocity of each car in this scenario?

The velocity of each car can be calculated by dividing the distance traveled by the time it takes to pass each other. This will give the velocity in meters per second or miles per hour.

What factors can affect the outcome of two cars passing each other with constant velocity?

The outcome can be affected by factors such as the speed of the cars, the distance between them, and the width of the road. Other factors may include the presence of obstacles or road conditions.

Can two cars pass each other with constant velocity in a curved road?

Yes, two cars can pass each other with constant velocity in a curved road as long as they maintain a constant speed and follow the curvature of the road. However, the velocity will not be constant in terms of direction, as it will change due to the curvature of the road.

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