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anonymous

  • one year ago

In the figure here, a red car and a green car 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 = 223 m. If the red car has a constant velocity of 20.0 km/h, the cars pass each other at x = 43.1 m. On the other hand, if the red car has a constant velocity of 40.0 km/h, they pass each other at x = 76.9 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

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  1. anonymous
    • one year ago
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    \[V^2 = V^2 +2a(x-xo)\]

  2. rajat97
    • one year ago
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    i don't know the formula but i'm sure that we need to use the concept of relative motion in this question i'll show you how

  3. rajat97
    • one year ago
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    just wait for some time and i'll get it to you

  4. anonymous
    • one year ago
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  5. anonymous
    • one year ago
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    I'm planning to use the distance travelled of the green car

  6. anonymous
    • one year ago
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    so I can use the formula which I don't know if it work.

  7. rajat97
    • one year ago
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    look the formulas that you posted is for simple linear uniformly accelerated motion here the easiest way to solve the question will be by using the concept of relative motion we cannot use the above formulas directly in relative motion we need to alter them a bit well, do you know relative motion??

  8. anonymous
    • one year ago
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    relative motion I see. I think we haven't tackled that topic but it will be a great advantage if I can study it now!

  9. rajat97
    • one year ago
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    yup you are a cool curious buddy i see! but we will first solve it without using relative motion and then we will jump to the topic as you have not yet studied it and you should be able to solve it without relative motion

  10. anonymous
    • one year ago
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    Yeaaaa I assume we can solved it

  11. anonymous
    • one year ago
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    http://imgur.com/zw7Qpbw

  12. anonymous
    • one year ago
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  13. anonymous
    • one year ago
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    the illustration of the question

  14. rajat97
    • one year ago
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    here in the first case the red car has a constant speed(20km/h) and no acceleretion and it travels 43.1m i.e. 0.0431km and meet the green car so the time elapsed will be = distance/speed = .0431/20=0.002155hrs(you can convert it to any other unit if you want) let the acceleration of the green car be a and it's initial speed be v it has travelled a distance of 233-43.1=189.9m=0.1899km what's the formula that relates the the time for which it travels, initial velocity, the acceleration, of the object and the distance or more correctly the displacement of an object is...... you know it it is \[s=ut + (1/2) at^2\] where s is the displacement, u is the initial velocity, a is the acceleration of the object and t is the time for which it travels now, we have s=0.1899, we need to find u and a and t=0.002155hrs now in the second case, the red car has a constant speed of 40km/h and all the things with the green car are the same except the time and the displacement 's' now, the red car travels 76.9m=0.0769 km so the time elapsed = 0.0769/40 = 0.0019225hrs now you just need to put s=233-76.9=156.1=0.1561km and t=0.0019225hrs now you have two equations(from the first and the second case) and two variables(u and a) just solve those two equations and you'll get the answer, albeit the numbers to be input in the equations are horrible

  15. rajat97
    • one year ago
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    i may not be very clear so feel free to ask anything from the answer:) and once you get it we will move towards relative motion

  16. rajat97
    • one year ago
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    and thanks for the medal:)

  17. anonymous
    • one year ago
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    Will study your text. Brb

  18. anonymous
    • one year ago
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    attempting to answer

  19. rajat97
    • one year ago
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    i'll be back by 20:00 (IST) gotta go now

  20. anonymous
    • one year ago
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    0.1461 = u(0.0019225) + (1/2)a(0.0019225)^2 I bring down the equation except u to the bottom right hand

  21. anonymous
    • one year ago
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    (0.1461-(1/2)a(0.0019225)^2)/(0.0019225) = u

  22. anonymous
    • one year ago
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    I think I got the wrong answer for getting a =64391 lol wut happened

  23. anonymous
    • one year ago
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  24. anonymous
    • one year ago
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    the numbers are so tedious maybe ill just convert km/h to m/s

  25. anonymous
    • one year ago
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    I got 5.06m/s^2 and 3.60m/s

  26. anonymous
    • one year ago
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    20km/h = 5.56 m/s 40km/h = 11.11 m/s then proceed to do the steps

  27. rajat97
    • one year ago
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    thanks for fanning me you are right we need to convert the values to S.I. units to solve the problem FIRST CASE: first of all we convert the speeds of the car as you have done and it comes out to be the same as you have posted next, we convert time into seconds we get, the time in the first case = t1(say) = 0.002155x60x60 = 7.758s Now, we convert the distance travelled by the green car into meters the distance = x1(say) = 189.9m so we the equation that we get from the first case is, 189.9 = 7.758u + (1/2)a(7.758)^2 so finally, EQATION FROM FIRST CASE: -------------------------------------------------------------------------- 7.758u + (30.093282)a = 189.9 ---- equation one -------------------------------------------------------------------------- now, we do it for the second case SECOND CASE: The time in the second case = t2(say) = 6.921s and the distance travelled by the green car = 156.1m so from the second case we get the equation 156.1 = 6.921u + (1/2)a(6.921)^2 finally, the equation from the second case is: -------------------------------------------------------------------------- 6.921u + (23.9501205) = 156.1 ----equation two -------------------------------------------------------------------------- now when we solve these two equations for 'a' and 'u' , we get a=4.59612 m/s^2 and v=6.64965m/s now you can convert the values to get the answers in km/h^2 and km/h hope this helps you!

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