velocity problem

Discussion in 'Physics' started by braddy, Aug 31, 2005.

  1. braddy

    Thread Starter Well-Known Member

    Dec 29, 2004
    83
    0
    Hi,
    i don't know how to calculate the velocity.
    The problem is:
    Use the velocity of a particle as a function of time tabulated below , to calculate the position of the particle at eact of the times ( assume at t=0 ,the particle is at the origin)

    Times(s)
    0.0
    .5
    1.5
    2.5
    3.5

    velocity (m/s)
    0.00
    0.75
    1.75
    8.75
    21.75

    position (m)
    0.00
    0.19
    1.44
    6.69
    21.94


    The answer is the column in red, BUT I don't know how to find those values.
    please Can someone can help me how to find the values of the positions.

    I tried the formula x=v*t but it does not work.

    please help

    Thank you

    Bertrand
     
  2. Haus76

    New Member

    Aug 31, 2005
    7
    0


    from the tables psted it appears to have an acceleration compontent. X=V*T only works for constant velocity. at this time i cannot recall the correct formula.
     
  3. Raspider

    New Member

    Jan 31, 2007
    1
    0
    Hello Brady, this problem is rather tricky. The Acceleration is not constant, it is changing therefore your formula x=V*T cannot be correct as it assumes constant acceleration.

    Therefore you must use the following formula. By the (0) I mean naught or X2 if you will.

    x-x(0) = (V^2 -V(0)^2)/2a)

    So for instance assume the particle starts at x=0 V=0 Y=0

    At your first time .5s the velocity is .75 M/S. Let's caculate the acceleration which is just the Change in Velocity/ Change in Time. So .75M/S-0M/S / .5M/S-0M/S which gives me 1.5=A

    So I plug it into the original formula. (.75M/S)^2 - (0M/S)^ / 1.5(2) which in turn gives me .1875 which they rounded to .19

    On the second step you will have to add .19 to the solution you obtain from step 1 to obtain X. A is 1/1 = 1

    (1.75M/S)^2-(.75M/S)^2 / 2(A) = 1.25 + .19 = 1.44 which is now X1.

    and so on and so forth.
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    i believe the above problem can be also solved with the help of graph
    plot graph of vel vs time, the area under the curve gives displ
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    The acceleration is given by V2 (velocity sub 2) - V1 over T2 - T1, where the velocity sub 2 is the velocity at time 2, and so on.

    The velocity at any time is given by the formula - V = V0 (V sub zero) + AT (acceleration times time).
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    it might be very tedious to solve it analytically as
    v = u + at
    the above formula assumes constant acceleration.
    and integration can not be applied since change of acceln as a functn of
    time is not known
    trying graphically might help
     
  7. noodle

    New Member

    Sep 14, 2007
    5
    0
    In this case the acceleration is not constant. So break it up into small intervals and apply equation s=ut+1/2at^2, which gives the distance travelled.
    Ill illustrate for first two intervals(1) t varies from 0 to 0.5 secs and velocity varies from 0 to 0.75 m/s^2.
    Therfore acceleration is change in velocity/time taken a=0.75/0.5=1.5m/s^2
    Applying equation s=0+1/2*1.5*0.5^2=0.19m.
    (2)t varies from 0.5 to 1.5 secs and velocity varies from 0.75 to 1.75m/s^2
    a=1m/s^2.
    distance travelled s=0.75*1+1/2*1*1^2=1.25m.
    Initially the position of the particle is 0.19m. therfore new position is 1.25+0.19=1.44m. similarly do the rest.:)
     
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