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A ball is dropped from a state of rest at time t=0.The distance traveled after t seconds is s(t)=16t^2 ft.
(a) How far does the ball travel during the time interval [3,3.5] ?Δs=____ft(b) Compute the average velocity over [3,3.5] .Δs/Δt= ____ ft/sec(c) Compute the average velocity over time intervals [3, 3.01] , [3, 3.001] , [3, 3.0001] , [2.9999, 3] , [2.999, 3] , [2.99, 3] .
Use this to estimate the object's instantaneous velocity at t=3 .V(3)= ____ ft/sec

Respuesta :

Answer:

i) Δs  = 196 - 144 = 52 feet

ii) Δs / Δt = 104 feet/second

iii) Therefore the velocity at t = 3 is v(3) = 96 feet/second

Step-by-step explanation:

a.) i) s = [tex]16t^{2}[/tex]

   ii) at 3 seconds the distance traveled [tex]s(3) = 16\times (3)^2\hspace{0.2cm} = 16 \times 9 = 144 feet[/tex]

   iii) at 3.5 seconds the distance traveled [tex]s(3.5) = 16\times (3.5)^2\hspace{0.2cm} = 16 \times 12.25 = 196 feet[/tex]

  iv) Δs  = 196 - 144 = 52 feet

b) Average velocity over [3, 3.5] = Δs / Δt  = 52/(0.5) = 104 feet/second

c) the average velocity over the interval

  I) i) [3, 3.01]

     ii) [tex]s(3.01) = 16\times (3.01)^2\hspace{0.2cm} = 16 \times 9.0601 = 144.962 feet[/tex]

     iii) Δs  = 144.962 - 144 = 0.962 feet

     iv) Average velocity over [3, 3.01] = Δs / Δt  = 0.962/(0.01) =

          96.2 feet/second

II) i) [3, 3.001]

     ii) [tex]s(3.001) = 16\times (3.001)^2\hspace{0.2cm} = 16 \times 9.006001 = 144.096 \hspace{0.1cm} feet[/tex]

     iii) Δs  = 144.096 - 144 = 0.096 feet

     iv) Average velocity over [3, 3.001] = Δs / Δt  = 0.096/(0.001) =

         96 feet/second

III) i) [3, 3.0001]

     ii) [tex]s(3.0001) = 16\times (3.0001)^2\hspace{0.2cm} = 16 \times 9.00060001 = 144.0096 \hspace{0.1cm} feet[/tex]

     iii) Δs  = 144.0096 - 144 = 0.0096 feet

     iv) Average velocity over [3, 3.0001] = Δs / Δt  = 0.0096/(0.0001) =

         96 feet/second

IV) i) [2.9999, 3]

     ii) [tex]s(2.9999) = 16\times (2.9999)^2\hspace{0.2cm} = 16 \times 8.9994 = 143.99\hspace{0.1cm} feet[/tex]

     iii) Δs  = 144 - 143.99 = 0.0096 feet

     iv) Average velocity over [2.9999, 3] = Δs / Δt  = 0.0096/(0.0001) =

         96 feet/second

V) i) [2.999, 3]

     ii) [tex]s(2.999) = 16\times (2.999)^2\hspace{0.2cm} = 16 \times 8.994 = 143.904\hspace{0.1cm} feet[/tex]

     iii) Δs  = 144 - 143.904 = 0.09598 feet

     iv) Average velocity over [2.999, 3] = Δs / Δt  = 0.09598/(0.001) =

         95.98 feet/second

VI) i) [2.99, 3]

     ii) [tex]s(2.99) = 16\times (2.99)^2\hspace{0.2cm} = 16 \times 8.94 = 143.0416\hspace{0.1cm} feet[/tex]

     iii) Δs  = 144 - 143.042 = 0.098 feet

     iv) Average velocity over [2.99, 3] = Δs / Δt  = 0.098/(0.01) =

         98 feet/second

 Therefore the velocity at t = 3 is v(3) = 96 feet/second