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kinematics_AP Physics 1

2022-09-16 00:01 作者:第一性原理  | 我要投稿

知识点

例题

解题过程

  1. GIVEN:

    v0=20m/s (upward)

    vf=0m/s

    FIND: t=?

    SOLVE:

    by formula: vf=v0+at

    0=20+9.8t

    t=2.04s

  2. GIVEN:

    v0=150m/s

    vf=0m/s (stop)

    a=-3m/s^2

    FIND:  delta x=?

    SOLVE:

    by formula: vf^2=v0^2+2a delta x

    0^2=150^2+2(-3)delta x

    delta x= 3750 m


  3. GIVEN:

    v0=-25m/s

    t=2s

    FIND: vf=?

    SOLVE:

    by formula: vf=v0+at

     vf=-25+(-9.8)*2=-44.6m/s


  4. GIVEN:

    v0=0m/s (accelerate from rest)

    a=25m/s^2 (upward)

    t=5s

    FIND: %5CDelta%20x

    SOLVE:

    %5CDelta%20x=%5CDelta%20x1+%5CDelta%20x2

    Before 5 s of accelerating

    by formula:  %5CDelta%20x=v0*t+(1/2)a*t^2

     %5CDelta%20x1=v0*t+1/2at^2=0+(1/2)*25*5^2=312.5 m

    v0'=v0+at=0+25*5=125 (continue to accelerate upward)

     to FIND: t

    vf=0 ( no accelerat)

    vf=v0+at

    0=125+(-9.8)t

    t=12.76 s

    After 5 s of accelerating

     %5CDelta%20x2=v0*t+1/2at^2=125*12.76+(1/2)(-9.8)(12.76^2)=797.2 m 

    %5CDelta%20x=%5CDelta%20x1+%5CDelta%20x2=312.5+797.2=1109.7 m


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