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Kinematics problems

  1. Sep 23, 2004 #1
    Hey all, I'm in university right now, though these questions are more geared towards high school. I've completed junior and senior physics in high school, and for some reason I am completely stumped on these two questions.

    1. A speeding car passes a highway patrol checkpoint and then accelerates at a constant rate. 4.05 s later, the car is 240 m from the checkpoint and its speed is then 19.7 m/s. What is the acceleration of the car?

    Since they don't mention V1 (initial) I went ahead and tried the equation
    [itex] d= v1t + 1/2(a)(t)^2 [/itex]

    I assumed that V1 was 0, though I don't believe it is since it mentions that the car was speeding when it passed the checkpoint. This left me with (solving for acceleration)

    [itex] d= 1/2 a(t)^2 [/itex]
    Which worked out to 29.26 m/s^2. I know this is wrong due to the fact that i need to submit my answer online and it shows that this is an incorrect answer.

    The next problem i had difficulty with:

    A ball is thrown straight up from the surface of the earth with and initial speed of 19.2 m/s. Neglecting any effects due to air resistance, what is the magnitude of the ball's displacement (from the starting point) after 0.70 s has elapsed?

    t = 0.70 s
    V1 = 19.2 m/s
    a = -9.8 m/s^2

    Therefore I assumed i could easily use [itex] d= v1t + 1/2at^2 [/itex]

    This works out to 15.84 m, which seems like a resonable answer, though it was wrong. I've also tried entering -15.84, but no luck.

    Any help you guys can offer is greatly appreciated!
  2. jcsd
  3. Sep 23, 2004 #2
    For the first problem, you can't just assume the initial velocity is zero. You have to calculated it using the formula: d=average velocity x time. Once you have solved for V1, you can use the formula: a=average velocity/time.

    As for the second problem, I am pretty sure you have the right answer. I would have done the same thing you did. :smile:
  4. Sep 23, 2004 #3
    for ur second problem with the ball i get 11.039m

    D= (19.2*.7)+(.5*-9.8*.7^2)

    As for the first one......i used this method (might not be valid)
    av vel=D/T
    Vo=98.82 m/s

    D=1/2(Vf+Vo)T...used this to check my Vo answer (alittle off b/c of rounding)

    240= (98.82*4.05)+(.5*a*4.05^2)
    a=-19.54 m/s^2 (of course not "exact" b/c of rounding...u may need to fix that in ur final answer b4 u submit it)

    BTW the car's Vo of 98.82 m/s is 221.4 mph ;)...freakin F1 racing and final vel of 19.7 m/s is only 44.1 mph

    because of this i think my answer is up for question :(
  5. Sep 23, 2004 #4
    EDIT: Thanks for the help guys. Your answer for the ball question is correct, silly me forgot a negative sign.

    As for the checkpoint question the answer doesn't seem to work unfortunately. I'm gonna look over it some more and try to figure it out.

    Your help is greatly appreciated.

    EDIT #2: We posted at the same time :p . I replied before seeing yours. Thanks again
    Last edited: Sep 23, 2004
  6. Sep 23, 2004 #5
    ur welcome......that checkpoint prob. is tricky ;)
    Last edited: Sep 23, 2004
  7. Sep 24, 2004 #6
    Yet other problem:

    A motorist travelling at 97 km/hr is being chased by a police car at 127 km/hr. If the police car starts from 2.0 km back, how long does it take to catch the motorist? Leave time in hours.

    I've tried everything i can think of to solve this problem. Can anyone clue me in here?
  8. Sep 24, 2004 #7
    Try this : the motor : x = 2 + 97t
    for the car : x = 127t

    Set these to equal and solve for t.the time is expressed in hours here !!!

  9. Sep 25, 2004 #8
    Cool Marlon, that worked perfectly, thanks a lot.

    I was also wondering if anyone could help me with this.

    An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.

    Time (s) Position, (m)
    47.50 9.000
    48.90 14.152
    50.30 22.048

    Calculate the magnitude of the acceleration at t=48.90 s.

    I've tried everything i can think of for this, but it just does not want to work.
    Any help is greatly appreciated
  10. Sep 25, 2004 #9
    The general equation is x = x_i + v_i * t + at²/2 where the _i denotes the initial position and velocity. You have three unknowns (x_i,v_i,a), yet you can set up three equations when you substitute x and by the corresponding given values. just solve them three equations in order to find the three unkowns and your problem is solved...

  11. Sep 25, 2004 #10
    Your reply seems to be over my head. Why exactly is x_i unknown? Doesn't it just correspond to distance?

    What I attempted to do to solve this problem was find the change in velocity between point a and b, then b and c. Getting the change in velocity between a and b would give me the velocity for the middle of those points, as would b and c. In order to find point b i would assume that i would get the average of these 2 points basically, though that doesn't seem to work. Understand what i mean?
  12. Sep 25, 2004 #11
    no, x_i is the initial distance from the y-axis at which the object started to move. You see, it is not always the case that an object starts moving from the origin.

    Let's say a wall is the origin. Suppose you stand against the wall (so you are at the origin) and i am 5 meters in front of you. this means that whenever we start to move i started 5 meters before you so your x_i = 0 and mine x_i = 5


    you know how to solve three equations with three unkowns ??? just asking
  13. Sep 25, 2004 #12
    Sorry bro, I understand what you're talking about with x_i, displacement basically. I'm just incredibly confused. I'll keep trying at it, thanks anyways though!
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