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Velocity and Acceleration of Cars

  1. Sep 1, 2008 #1
    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 = 230 m. If the red car has a constant velocity of 24.0 km/h, the cars pass each other at x = 45.5 m, and if it has a constant velocity of 43.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the acceleration of the green car?

    I know that if xr = 45.5m then xg = 184.5m. And if xr = 76.6m then xg = 153.4m.

    I'm not sure what to do with these numbers. Where do I go from here?
  2. jcsd
  3. Sep 1, 2008 #2
    You attempt the problem.
  4. Sep 1, 2008 #3
    That's the thing. I don't really know how.
    What is s and u in that equation? I don't really understand what you said in your explanation in that post.
  5. Sep 2, 2008 #4
    Well let's say that the green car starts at x = 0. Then s1 = -184.5m and s2 = -153.4m for the two cases.. it is the distance travelled by the green car.

    Now since the red car travels at constant velocity we can say how long the red car was travelling for up to crossing... i.e.

    t_1 = 45.5 / k(24.0) \quad & \quad t_2 = 76.6/k(43.0)

    where k is just a conversion constant to SI units becuase the speeds are in kilometers per hour and you want to work in m/s.

    Now you have two equations for the green car:

    s_1 = ut_1 + \frac{1}{2}a{t_1}^2
    s_2 = ut_2 + \frac{1}{2}a{t_2}^2

    where u is the initial velocity and a is the acceleration -- precisely what you're trying to find.

    Now you know t_1 and t_2 corresponding to s_1 and s_2 (since the red car and green car will travel for the same length of time when they cross)

    So you just have two simultaneous equations.
  6. Sep 2, 2008 #5
    s in this case is displacement (should be x since you mentioned the x-axis), and u is velocity (usually written as a v) with a, and t representing acceleration and time respectively. The key to the problem is determining the times involved. For instance, at 24.0 km/h, the red car travels 45.5 m. How long did it take to get there (remember unit conversions)? That is the same time it took the green car to travel a distance 45.5 - 230 m = -184.5 m (negative if you've set the positive x-axis to be the direction the red car is traveling. This alone isn't enough to solve for the acceleration AND initial velocity, so the second data point is needed. A comparison between your two sets of answers should yield the results if you use the kinematic equation you described.

    Note: If you set the problem up the way I have described: (red car traveling in the positive x-direction), your initial acceleration and velocity for the green car will be negative to signify motion in the negative x-axis.
  7. Sep 2, 2008 #6
    Ok so I plug in all the numbers I have. However, I end up having two equations with two unknowns. The unknowns being the things you need to find. Do I have to assume one of the values for one of the equations? Or do I solve through a system of equations?
  8. Sep 2, 2008 #7
    \frac{s_1}{t_1} = u + \frac{1}{2}at_1
    \frac{s_2}{t_2} = u + \frac{1}{2}at_2

    can't you get rid of u and then find a?
  9. Sep 2, 2008 #8
    If I did that, I could get rid of u.
    Then that would leave me with:

    [tex](s_1/t_1) - (s_2/t_2) = (\frac{1}{2}at_1) - (\frac{1}{2}at_2)[/tex]

    Shouldn't [tex]t_1[/tex] and [tex]t_2[/tex] in [tex](\frac{1}{2}at_1) - (\frac{1}{2}at_2)[/tex] be squared?

    I find out that a is -15.11 m/s^2 from all of this. Does that sound right? Why do I get a negative acceleration?
    If I have it squared, my a is -1.141 m/s^2.
  10. Sep 2, 2008 #9
    They shouldn't be squared because I divided through by t to leave a single u in each equation.

    The green car starts 230m in front of the red car and ends up going backwards to the points where the cars cross.. in the equations you see that it travels a negative distance (s is negative) so a ends up negative...

    Deklan said that u and a will both be negative, but it is possible that the green car is initially travelling forwards and then decelerates (in which case the velocity will decrease until it is negative)... i'm not sure what value you'll get for u when you plug in for a, t and s...
  11. Sep 2, 2008 #10
    ok the next question is do I plug a into the original equation? or the modified equation that we used to find a?
  12. Sep 2, 2008 #11
    You can sub. for a in either equation in post #4.
    Where before you knew only 2 variables, now you know 3 and have only 1 unknown, that unknown being u.
    A bit of algebra and bob's your uncle.
    The equations in post #7 are the same as in #4 you should realise, just rearranged, so you might think it easier to use those...
  13. Sep 2, 2008 #12
    Ok if I plug in a for the first original equation in post #4, u = 24.53665 m/s

    I tried the answer. It was incorrect.
  14. Sep 2, 2008 #13
    I do not get that value for u...

    (s - 0.5at^2)/t = u

    s is negative and quite large, a is negative but |s| > 0.5at^2 so you're going to get negative u for a start.
    I actually get -23.15m/s using eq'n. 1.
  15. Sep 2, 2008 #14
    I'm off to bed, hopefully we've cracked this problem and just need to review some of the plugging in of numbers.
  16. Sep 2, 2008 #15
    Your answer is incorrect as well.
    What numbers did you use? I keep getting 24.53 as my answer all the time.
    Last edited: Sep 2, 2008
  17. Sep 3, 2008 #16
    I have attempted this problem on my own on paper now. Before I was just thinking out loud I suppose. I went ahead and switched the direction of our coordinate axis so that the GREEN car moves in the positive x-direction, beginning at x=0, and the RED car is now at x=230m and moving to the left (negative direction) hopefully this will clear up the sign issue.

    I achieved the results of initial velocity v=-23.1 m/s (meaning it starts off moving in the SAME direction as the RED car, but then turns around because of it's positive acceleration of 14.3 m/s^2

    Here's how I did it, and I could very well be wrong.

    To me, it seems that the information about the red car is only there to infer information of the green car. For instance, at 43.0 km/h it intersects the green car at 76.6 m right? So it takes a time of 6.41 s for the green car to travel 153.4 m. Does that make sense? Similarly at 24.0 km/h they intersect at 45.5 m, so it takes the green car 6.83 s to travel 184.5 m. So now we have relevant data for the green car...

    x1 = 153.4 m
    x2 = 184.5 m
    t1 = 6.41 s
    t2 = 6.83 s

    Where x1 is the distance the car travels in a time t1, and x2 is the distance the car travels in a time t2.

    We now use some kinematics. See the attached .doc file:


    Notice that that units all work out correctly in this sequence (that velocity is indeed in m/s).

    You can now use this initial velocity to solve for the acceleration, which I get to be 14.7 m/s^2.

    I see that Looney got a similar result, but you said it was wrong. My work looks right, and the units work out perfectly, so I don't know what to tell you... I could be wrong. The fact that the velocity is negative is a little precarious, but not outside the realm of what we are dealing with here.

    Hope this helps.

    Attached Files:

  18. Sep 3, 2008 #17
    That math seems right. However, as I tried to put in answers for my homework, the correct acceleration of the green car is -15.11 m/s^2.
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