When a ball is dropped from a state of rest at time t=0t=0, the distance, measured in feet, that it has traveled after tt seconds is given by the formula s(t)=16t2s(t)=16t2 . (use decimal notation. if necessary, give your answer rounded to two decimal places.) (a) how far does the ball travel during the time interval [3, 3.5 ]? distance = .5 feet (b) compute the average velocity over the time interval [3, 3.5 ]. average velocity = 104 feet/sec (c) by computing the average velocity over the time intervals [3, 3.1 ], [3, 3.01 ], and [3, 3.001 ], . . , estimate the ball's instantaneous velocity at t=3t=3. instantaneous velocity = feet/sec?

Part A. To solve for the distance travelled during the interval, all we have to do is to plug in values of t = 3 and t = 3.5 in the equation and the difference would be the

when t = 3: s = 16 (3)^2 = 144 m

when t = 3.5: s = 16 (3.5)^2 = 196 m

Therefore the distance travelled within the interval is:

196 m – 144 m = 52 m

Part B. The velocity is calculated by taking the 1st derivative of the equation. v = ds / dt

s = 16 t^2

ds / dt = 32 t = v

when t = 3: v = 32 (3) = 96 m / s

when t = 3.5: v = 32 (3.5) = 112 m / s

Therefore the average velocity is:

(96 + 112) /2 = 104 m / s

Part C. We can still use the formula v = 32 t and plug in the value of t = 3

v = 32 t = 32 (3)

v = 96 m / s

2 is the final ! : )

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