? asked in 科學及數學數學 · 1 decade ago

difficult differentiation ex

Update:

The answers are 6m/s,36and27m/s respectively.

4 Answers

Rating
  • 1 decade ago
    Favorite Answer

    Given that the displacement of P at time t is s=15t+6t²-t³

    Let f(t) be the s at time t,so

    f(t)=15t+6t²-t³

    As f(0)=0, s=0 when t=0,

    the average velocity over the first t seconds equals s/t

    s/t=f(2)/2

    s/t=[15(2)+6(2)²-(2)³]/2

    s/t=23

    Thefore, the average velocity of P over the first 2 seconds is 23m/s

    (b)

    Given that the displacement of P at time t equals s=15t+6t²-t³

    From (a), f(t)=s=15t+6t²-t³

    ds/dt=v=f ’(t)=d(15t+6t²-t³)/dt

    ds/dt=v=f ’(t)=-3t²+12t+15

    When v=0, P comes to instantaneous rest

    v=0

    -3t²+12t+15=0

    (t+1)(t-5)=0

    t=-1 or t=5

    As t=-1 is meaningless, P comes to instantaneous rest at t=5

    Therefore, when P comes to instantaneous rest,

    s=f(5)

    s=15(5)+6(5)²-(5)³

    s=100

    The value of s when P comes to instantaneous rest is 100

    (c)

    From (a) and (b),

    s=f(t)=15t+6t²-t³

    v=f ’(t)=3t²+12t+15

    dv/dt=a=f"(t)=d(-3t²+12t+15)/dt

    dv/dt=a=f"(t)=-6t+12

    When the acceleration of P is instantaneously zero, a=0

    a=0

    -6t+12=0

    t=2

    Therefore, the acceleration of P is instantaneously zero at t=2

    v=f ’(2)

    v=-3(2)²+12(2)+15

    v=27

    The velocity of P when its acceleration is instantaneously zero is 27m/s

    • Commenter avatarLogin to reply the answers
  • Anonymous
    1 decade ago

    my solution: http://hk.geocities.com/samfai_king/solution.jpg

    希望可以幫到你

    2008-05-25 19:21:17 補充:

    http://hk.geocities.com/samfai_king/image1才對

    2008-05-25 19:22:33 補充:

    sorry, 應是http://hk.geocities.com/samfai_king/image-1.jpg才對

    • Commenter avatarLogin to reply the answers
  • wy
    Lv 7
    1 decade ago

    Since s=15t +6t^2 -t^3.

    Therefore, ds/dt= speed = 15 +12t -3t^2.

    when t=0, ds/dt=15, when t=2, ds/dt=27. herefore, average speed for the first 2 sec.=(15 +27)/2=42/2=21.

    When ds/dt=0, that is 15+12t-3t^2=0, t=5 or -1(reject). Therefore, at t=5,

    s= 15 x5 + 6 x 5 x 5 - 5 x5 x5 =100.

    Since speed, v=15 +12t -3t^2, therefore, accelaration=dv/dt=12 -6t=a.

    When a=0, t=2, therefore, v=15 + 12 x2 - 3 x2 x2=27.

    • Commenter avatarLogin to reply the answers
  • 1 decade ago

    (a)s=15t+6t²-t³

    v=ds/dt=15+12t-3t²

    t=0, v=15, t=1, v=24, t=2, v=27

    ∴Average speed=(15+24+27)/3=22metres/s

    (b)When P comes to instantanteous rest, v=0

    ∴15+12t-3t²=0

    t²-4t-5=0

    (t-5)(t+1)=0

    t=5 or -1(rej.)

    Value of s when t=5 is 75+150-125=100metres

    (c)a=dv/dt=12-6t

    ∴12-6t=0

    t=2

    Value of s what t=2 is 30+24-8=46metres

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.