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25 October, 20:51

The position function of an object moving along a straight line is given by s=f (t). The average velocity of an object over the time interval [a, b] is the average rate of change of f over [a, b]; it's velocity at t=a is the rate of change of f at a.

A ball is thrown straight up with an initial velocity of 144ft/sec, so that it's height (in feet) after t sec is given by s=f (t) = 144t-16t^2.

(A) What is the average velocity of the ball over the following time intervals?

[3,4] = ft/sec

[3,3.5] = ft/sec

[3,3.1] = ft/sec

(B) what is the instantaneous velocity at time t=3? Ft/sec

(C) what is the instantaneous velocity at time t=8? Ft/sec

(D) when will the ball hit the ground?

+5
Answers (1)
  1. 25 October, 20:57
    0
    A) Position at time t = 3 s:

    s₁ = 144*3 - 16*3² = 288 ft

    Position at time t = 4 s:

    s₂ = 144*4 - 16*4² = 320 ft

    Δs = s₂ - s₁ = 320 - 288 = 32ft

    Δt = t₂ - t₁ = 4 - 3 = 1 s

    Vavg[3, 4] = Δs / Δt = 32 / 1 = 32 ft/s

    Vavg[3, 3.5] = [ (144*3.5 - 16*3.5²) - (144*3 - 16*3²) ] / [3.5 - 3] = 40 ft/s

    Vavg[4, 4.1] = [ (144*3.1 - 16*3.1²) - (144*3 - 16*3²) ] / [3.1 - 3] = 4.64 ft/s

    B) lim [ (144t - 16t²) - (144*4 - 16*3²)) / (t - 3) ] =

    t→3

    lim [ (-16t² + 144t - 288) / (t - 3) ] =

    t→3

    48 ft/s

    C) lim [ (144t - 16t²) - (144*8 - 16*8²)) / (t - 8) ] =

    t→8

    lim [ (-16t² + 144t - 128) / (t - 8) ] =

    t→8

    lim (-16t + 16) =

    t→8

    -112 ft/s

    D) 144t - 16t² = 0

    (144 - 16t) t = 0

    t = 0

    or

    144 - 16t = 0

    t = 9
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