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Uniform acceleration

  1. Nov 7, 2005 #1
    :smile: Hi, I'm new here and I was really hoping someone would beable to help me ...
    Well theres two questions basically and I've literally been trying for hours to work them out. Well here goes:

    Two cars are travelling towards each other with equal speeds of 35m/s. When they are 500m apart they both decide to brake.
    a) what minimum equal decelerations would they require just to avoid an accident?
    The brakes on one car fail and it continues with the same speed. The other car slows down and at the point of collision it has just stopped
    b) what distances have the two cars travelled when the collision occurs?
    c) what time has elapsed?

    :uhh: are you ment to use simultaneous equations?

    The other one is:

    A police car stationary on the side of the road sees a car passing at 40m/s. the police car immediately gives chase and accelerates at 3m/s/s for 16s, followed by a constant speed.
    a) how long does it take for the police car to catch up the speeding motorist?
    b) what distance will the police car have travelled?

    Please help if you can - I challenge you! :blushing:
  2. jcsd
  3. Nov 7, 2005 #2
    Yes that's probably the better way to do it. Give it a shot.

    Solve the simultaneous equation to find the position where both equations are equal.

    Easy enough once you got the first one.
  4. Nov 7, 2005 #3
    Thanks so much for your quick response. Would you be able to help me further? I can't work out which equations to use in question 1b) ......
  5. Nov 7, 2005 #4
    If car one starts at x = 0 and car 2 starts at x = d, then they will meet at a point R between 0 and d. car 1 will have a constant velocity, car 2 will have a constant acceleration, and the distance they need to travel are R for car 1 and d-R for car 2.
  6. Nov 8, 2005 #5
    still stuck! anyone have any more advice or could thet let me know the answers they got?
  7. Nov 8, 2005 #6
    anyone pleeeeeeeeeaaaaaaase?
  8. Nov 8, 2005 #7
    Were you able to convert
    into 2 position equations?
  9. Nov 8, 2005 #8


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    Two equations for 1b, which can be solved via substitution:

    Look at the problem from the point of view of one of the cars. Their initial distance apart is 500 meters. The rate that the distance between the cars is decreasing is 70 meters/second (their coming towards each other at 35 m/s each). One car is decelerating, which is slowing the rate of decrease (in other words, it's opposite the closing velocity, or positive).

    Your second equation is the amount of time required to go from 35 m/s to 0. (v-at = 0). Rearrange to solve for either a or t. Substitute into the first equation and solve.

    Substitute the answer from your first equation into your second and solve the second variable.
  10. Nov 8, 2005 #9


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    Cop vs. speeder

    speeder d = vt
    Cop d= 1/2at^2

    what do you solve for? when do they meet?
  11. Nov 8, 2005 #10
    for 1b) i got 333.33m and 166.67m and 1c) a time of 9.5s.
    and for

    and for 2a) i got 13.33s and 2b) 533.33m

    what do you think?
    Last edited: Nov 8, 2005
  12. Nov 9, 2005 #11
  13. Nov 9, 2005 #12


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    Fabbo, your answer for 2a is completely wrong. You aren't getting the formula correct because you don't correctly interpret what is going on.

    The cop accelerates for a while, then proceeds at a constant speed. Obviously when the cop catches the car they will have driven the same distance, that's what catching means. So we have:

    Car_distance = Cop_accelerating_distance + Cop_constant_speed_distance

    Now fill in the formula and solve it.
  14. Nov 9, 2005 #13


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    Your answers for question 1 are correct, but I just wonder how you solved it. The question is put in a strange way, I would calculate the time first, the distances second. They ask for the distances first, but I don't see a nice way to find the distances without using the time.

    Just for informational value, I would have used:
    stopping_distance = 500 - driving_distance

    Filling in and solving this formula gives one the time as 9.52s, thereafter I would find the distances.
  15. Nov 9, 2005 #14
    try this!

    hey fabo...
    i will start of with the second probs first...just coz i m dead sure bout it.
    the police car is known ta hav an acceleration of 3/m/s/s...from the problem itself u cud get the idea that it was at halt...when it had ta start acceleratin ta chase the other car. so the initial velocity goes 0 rite there..final has ta be equal to the other car's so that it cud get the hang of them! final goes 40m/s ....now when we hav the Vi,Vf,a, not a big deal ta get time....Vf=Vi +at!
    now when u hav time put the info in the other equation ta get the distance!
    y= Vi t + .5 at^2....solve!!!
    the first one.....we hav the initial velocity of cars....knowin that they are aspirin ta stop....final velocity has ta be 0....distance that has ta be a limit within which they need ta stop is 500m. for acceleration.....use..VF^2= Vi^2+2ad....now that we know that the car is decceleratin at the speed of 1.2m/s/s....it stops rite at the moment of collision....to make it more clear...forget bout the part where it says,"where it collides....
    think through our equation...with the equation that we made...the cars wud stop with -1.2m/s/s (a) in 500 m....doesnt matter if the other ones still movin or not, the first car will "just "stop at 500m from its initial position. i wud still suggest u to search for the answer to the 1st question.....i cud just giv u some grounds ta think over!
    hope it helps u in someway.....happy physics...:tongue2:
  16. Nov 9, 2005 #15
    This is an incorrect solution. He would never reach the car but be following it forever.
  17. Nov 10, 2005 #16


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    doesn't the cop catch the speeder in 48 sec. both traveling 1920 meters or about 1.2miles, then writes him one of those $250.00 wreckless driving tickets for driving 90mph.
  18. Nov 11, 2005 #17


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    kp, I get a different answer.
  19. Aug 8, 2006 #18
    OK, what if..

    The trooper is sitting stationary at the side of the road, you pass by at 90MPH, he accelerates and catches you at the 3 mile (or 2 minute) mark.. or what if he catches you in 5 miles (3.3 minutes)

    How fast must/ can he accelerate, and how fast is he going when he catches you??
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