The problem provided asks us to find:

(a) The average velocity of the object during the first 4 seconds.

The height function is given as: $s(t) = -4.9t^2 + 160$ The average velocity $\overline{v}$ over a time interval $[0, 4]$ can be calculated using the formula: $\overline{v} = \frac{s(4) - s(0)}{4 - 0}$

Let's calculate this.

At $t = 0$: $s(0) = -4.9(0)^2 + 160 = 160 \, \text{m}$

At $t = 4$: $s(4) = -4.9(4)^2 + 160 = -4.9(16) + 160 = -78.4 + 160 = 81.6 \, \text{m}$

Now, the average velocity is: $\overline{v} = \frac{81.6 - 160}{4} = \frac{-78.4}{4} = -19.6 \, \text{m/s}$ Thus, the average velocity during the first 4 seconds is $-19.6 \, \text{m/s}$.

(b) Using the Mean Value Theorem to find when the instantaneous velocity equals the average velocity.

The instantaneous velocity is the derivative of the height function $s(t)$: $v(t) = \frac{d}{dt} \left( -4.9t^2 + 160 \right) = -9.8t$ According to the Mean Value Theorem, there exists a time $c \in (0, 4)$ such that: $v(c) = \overline{v}$ We already know that $\overline{v} = -19.6 \, \text{m/s}$, so: $-9.8c = -19.6$ Solving for $c$: $c = \frac{-19.6}{-9.8} = 2 \, \text{seconds}$ Thus, at $t = 2$ seconds, the instantaneous velocity equals the average velocity.

Would you like further details on these steps or have any questions?

Related Questions:

1. How is the Mean Value Theorem applied in different types of motion?
2. What is the significance of negative velocity in free fall problems?
3. How would the average velocity change if the time interval was extended?
4. Can the Mean Value Theorem be applied if the object reaches terminal velocity?
5. What happens to the object’s height function as time approaches infinity?

Tip:

In physics problems involving motion, carefully distinguishing between average and instantaneous quantities helps solve many real-world motion scenarios, such as free fall.