Question

A ball is dropped from a state of rest at time t=0.
The distance traveled after t seconds is s(t)=16t2 ft.

(a) How far does the ball travel during the time interval [6,6.5] ?
Δs= ___ ft

(b) Compute the average velocity over [6,6.5] .
Δs/Δt= ___ ft/sec

(c) Compute the average velocity over time intervals [6, 6.01] , [6, 6.001] , [6, 6.0001] , [5.9999, 6] , [5.999, 6] , [5.99, 6] .Use this to estimate the object's instantaneous velocity at t=6 .
V(6)= ____ ft/sec

Answer 1

a. s(t) = 4 ft

b. Average velocity = 8 ft/ sec

c. Average velocity = 400 ft/ sec

Average velocity  = 4000 ft/sec

Average velocity = 40000 ft/ sec

Average velocity = 400, 000 ft/ sec

Average velocity = 400 ft/ sec

How to determine the value

Given the expression;

s(t)=16t^2 ft

a. Time interval [6, 6. 5}

Time interval = 6. 5 - 6

Time interval = 0. 5

s(t) = 16 ( 0. 5)^2

s(t) = 16(0. 25)

s(t) = 4 ft

b. Average velocity is expressed as;

Average velocity = velocity/ time taken

Average velocity = 4/ 0. 5 = 8 ft/ sec

c. {6, 6. 01} , Interval = 6. 01 - 6 = 0. 01

Average velocity = 4/ 0. 01 = 400 ft/ sec

[6, 6.001], Interval = 6. 001 - 6 = 0.001

Average velocity = 4/ 0. 001 = 4000 ft/sec

[6, 6.0001], Interval = 6. 0001 - 6 = 0. 0001

Average velocity = 4/ 0. 0001 = 40000 ft/ sec

[5.9999, 6] , Interval = 6 - 5. 9999 = 0. 00001

Average velocity = 4/ 0. 00001 = 400, 000 ft/ sec

[5.99, 6] , Interval = 6 - 5. 99 = 0. 01

Average velocity = 4/ 0. 01 = 400 ft/ sec

Thus, the time interval is estimated by subtracting the initial value from the final value.

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