The deck of a bridge is suspended 295 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by
y = 295 − 16t2.
(a)
Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 3 and lasting the following amount of time.
(i)
0.1 seconds
ft/s
(ii)
0.05 seconds
ft/s
(iii)
0.01 seconds
ft/s
(b)
Estimate the instantaneous velocity (in ft/s) of the pebble after 3 seconds.
ft/s

Let's break down the problem step by step.

### Part (a): Average Velocity

The average velocity over a time interval \([t_1, t_2]\) is given by:

\[
\text{Average Velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

Given that the height \( y(t) \) of the pebble above the water surface after \( t \) seconds is described by the equation:

\[
y(t) = 295 - 16t^2
\]

We're asked to find the average velocity for three different time intervals, starting from \( t = 3 \) seconds.

#### (i) 0.1 seconds

The interval is from \( t_1 = 3 \) to \( t_2 = 3.1 \).

1. Calculate \( y(3) \):

\[
y(3) = 295 - 16(3)^2 = 295 - 144 = 151 \text{ feet}
\]

2. Calculate \( y(3.1) \):

\[
y(3.1) = 295 - 16(3.1)^2 = 295 - 16(9.61) = 295 - 153.76 = 141.24 \text{ feet}
\]

3. Calculate the average velocity:

\[
\text{Average Velocity} = \frac{141.24 - 151}{3.1 - 3} = \frac{-9.76}{0.1} = -97.6 \text{ ft/s}
\]

#### (ii) 0.05 seconds

The interval is from \( t_1 = 3 \) to \( t_2 = 3.05 \).

1. Calculate \( y(3.05) \):

\[
y(3.05) = 295 - 16(3.05)^2 = 295 - 16(9.3025) = 295 - 148.84 = 146.16 \text{ feet}
\]

2. Calculate the average velocity:

\[
\text{Average Velocity} = \frac{146.16 - 151}{3.05 - 3} = \frac{-4.84}{0.05} = -96.8 \text{ ft/s}
\]

#### (iii) 0.01 seconds

The interval is from \( t_1 = 3 \) to \( t_2 = 3.01 \).

1. Calculate \( y(3.01) \):

\[
y(3.01) = 295 - 16(3.01)^2 = 295 - 16(9.0601) = 295 - 144.9616 = 150.0384 \text{ feet}
\]

2. Calculate the average velocity:

\[
\text{Average Velocity} = \frac{150.0384 - 151}{3.01 - 3} = \frac{-0.9616}{0.01} = -96.16 \text{ ft/s}
\]

### Part (b): Instantaneous Velocity

The instantaneous velocity at \( t = 3 \) seconds can be found by taking the derivative of the height function \( y(t) \) and evaluating it at \( t = 3 \).

Given:

\[
y(t) = 295 - 16t^2
\]

The derivative with respect to \( t \) (which gives the velocity \( v(t) \)) is:

\[
v(t) = \frac{dy}{dt} = -32t
\]

Evaluate at \( t = 3 \):

\[
v(3) = -32(3) = -96 \text{ ft/s}
\]

### Final Answers:
- (a)
  - (i) The average velocity over 0.1 seconds is \(-97.6\) ft/s.
  - (ii) The average velocity over 0.05 seconds is \(-96.8\) ft/s.
  - (iii) The average velocity over 0.01 seconds is \(-96.16\) ft/s.
- (b) The instantaneous velocity at \( t = 3 \) seconds is \(-96\) ft/s.

Would you like more details or have any questions?

### Follow-up Questions:
1. What does the negative sign in the velocity indicate about the motion of the pebble?
2. How would the instantaneous velocity change if the initial height of the bridge were different?
3. Can you find the time when the pebble hits the water?
4. How does the average velocity compare to the instantaneous velocity as the time interval decreases?
5. What is the significance of finding the derivative in the context of instantaneous velocity?

### Tip:
The closer the time intervals used to calculate average velocity, the more accurately it approximates the instantaneous velocity at a specific moment.

Calculate the average and instantaneous velocity of a pebble falling from a bridge 295 feet above a river, given a height function y(t). Explore how to find average velocity over small time intervals and estimate instantaneous velocity at t = 3 seconds.

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