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15 January, 07:16

If a ball is thrown in the air with a velocity 52 ft/s, its height in feet t seconds later is given by y = 52t - 16t^2. Find the average velocity ofr the time period beginning when t = 2 and lasting 0.5 second. ft/s 0.1 second. ft/s 0.05 second. ft/s 0.01 second. ft/s Estimate the instantaneous velocity when t = 2. ft/s

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  1. 15 January, 09:28
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    a.) - 20ft/s

    b.) - 13.6ft/s

    c.) - 12.8ft/s

    d.) - 12.16ft/s

    e.) - 12ft/s

    Step-by-step explanation:

    Average Velocity = Change in distance/change in time.

    Distance from the question in given in form of t as y = 52t - 16t² If our initial time is 2, distance travelled at t=2 is given as

    Y (2) = 52 (2) - 16 (2) ² = 104 - 64

    Y = 40ft.

    For question a, when change In t is 0.5 seconds, that is from 2 sec to 2.5 seconds,

    Average velocity = y (2.5) - y (2) / 0.5

    y (2.5) = 52 (2.5) - 16 (2.5) ² = 130 - 100 = 30

    Y (2) = 40

    Average velocity in 0.5 seconds = [30 - 40]/0.5 = - 20ft/s.

    For question b, when change In t is 0.1 seconds, that is from 2 sec to 2.1 seconds,

    Average velocity = y (2.1) - y (2) / 0.1

    y (2.1) = 52 (2.1) - 16 (2.1) ² = 109.2 - 70.56 = 38.64

    Y (2) = 40,

    Average velocity in 0.1 seconds = [38.64 - 40]/0.1 = - 13.6ft/s.

    For question c, when change In t is 0.05 seconds, that is from 2 sec to 2.05 seconds,

    Average velocity = y (2.05) - y (2) / 0.05

    y (2.05) = 52 (2.05) - 16 (2.05) ² = 106.6 - 67.24 = 39.36

    Y (2) = 40

    Average velocity in 0.05 seconds = [39.36 - 40]/0.05 = - 12.8ft/s.

    For question d, when change In t is 0.01 seconds, that is from 2 sec to 2.01 seconds,

    Average velocity = y (2.01) - y (2) / 0.01

    y (2.01) = 52 (2.01) - 16 (2.01) ² = 104.52 - 64.6416 = 39.8784

    Y (2) = 40

    Average velocity in 0.5 seconds = [39.8784 - 40]/0.01 = - 12.16ft/s.

    Instantaneous velocity at t = 2 is derived by getting the first derivative of y and inserting Our value of t=2 into the first derivative.

    If y = 52t - 16t², then derivative of y becomes y' given as

    y' = 52 - 32t

    At t = 2,

    y' = 52 - 32 (2) = 52 - 64 = - 12ft/s.

    Instantaneous velocity at t=2 is given as - 12ft/s.
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