Question

When a ball is dropped from a state of rest at time t = 0, the distance it has traveled after t seconds is s(t) = 16t2. (a) How far does the ball travel during the time interval [4, 4.5]? 68 ft (b) Compute the average velocity over the time interval [4, 4.5]. ft/sec (c) By computing the average velocity over the time intervals [4, 4.100], [4, 4.010], and [4, 4.001], . . . , estimate the ball's instantaneous velocity at t = 4. (Round your answer to one decimal place.) ft/sec.

Answer 1

The ball travels 68 ft during the time interval [4, 4.5]. The average velocity over [4, 4.5] is 136 ft/s. The instantaneous velocity at t = 4 can be estimated by computing the average velocity over smaller and smaller time intervals approaching t = 4.

Step-by-step explanation:

(a) To find the distance the ball travels during the time interval [4, 4.5], we need to find the difference between the distance at time t = 4.5 and the distance at time t = 4. Using the equation s(t) = 16t^2, we can calculate:
Distance at t = 4.5: s(4.5) = 16(4.5)^2 = 324 ft
Distance at t = 4: s(4) = 16(4)^2 = 256 ft
The distance traveled during the time interval [4, 4.5] is the difference between these two distances: 324 ft - 256 ft = 68 ft.

(b) The average velocity over the time interval [4, 4.5] can be calculated by finding the difference in distance and dividing it by the difference in time. The average velocity is given by the formula:
Average velocity = (change in distance) / (change in time)
The change in distance is 68 ft (found in part (a)) and the change in time is 4.5 s - 4 s = 0.5 s. Therefore, the average velocity over the time interval [4, 4.5] is:
Average velocity = 68 ft / 0.5 s = 136 ft/s.

(c) To estimate the ball's instantaneous velocity at t = 4, we can compute the average velocity over smaller time intervals that approach t = 4. By using smaller and smaller time intervals, the average velocity will be a better approximation of the instantaneous velocity at t = 4. We can calculate the average velocity over the time intervals [4, 4.100], [4, 4.010], and [4, 4.001] as follows:

Average velocity over [4, 4.100]: (s(4.1) - s(4)) / (4.1 - 4)
Average velocity over [4, 4.010]: (s(4.01) - s(4)) / (4.01 - 4)
Average velocity over [4, 4.001]: (s(4.001) - s(4)) / (4.001 - 4)

By calculating these average velocities and getting smaller time intervals, we can estimate the ball's instantaneous velocity at t = 4.

Answer 2

Part a)

`d = 68 ft`

Part b)

`v_(avg) = 136 ft/s`

Part c)

`v = 128 ft/s`

Step-by-step explanation:

Part a)

As we know that displacement is given as

`s = 16 t^2`

now the displacement in 4.5 s and t = 4 s is given as

`y_1 = 16(4.5)^2 = 324 ft`

`y_2 = 16(4^2) = 256 ft`

now displacement is given as

`d = y_1 - y_2`

`d = 324 - 256`

`d = 68 ft`

Part b)

So displacement of the ball in given interval of t = 4.00 to 4.5 s

`d = 68 ft`

time interval is given as

`\Delta t = 0.5 s`

average velocity is given as

`v_(avg) = (68)/(0.5)`

`v_(avg) = 136 ft/s`

Part c)

So displacement of the ball in given interval of t = 4.00 to 4.1 s

`d = 16(4.1^2 - 4^2)`

`d = 12.96 ft`

average velocity is given as

`v_(avg) = (12.96)/(0.1)`

`v_(avg) = 129.6 ft/s`

Similarly for time interval t = 4.0 s to t = 4.01 s

`d = 16(4.01^2 - 4^2)`

`d = 1.2816 ft`

average velocity is given as

`v_(avg) = (1.2816)/(0.01)`

`v_(avg) = 128.16 ft/s`

Similarly for time interval t = 4.0 s to t = 4.001 s

`d = 16(4.001^2 - 4^2)`

`d = 0.128016 ft`

average velocity is given as

`v_(avg) = (0.128016)/(0.001)`

`v_(avg) = 128.016 ft/s`

so instantaneous velocity is given as

`v = 128 ft/s`