QUESTION TWO (15 MARKS)
(a) Consider a rock that is launched from the ground at a speed of . Neglecting air
resistance, a well known formula from physics states that the position of the rock after
seconds is given by the function . The position is measured in feet with
corresponding to the ground. Find the average velocity of the rock between each
pair of intervals given as
i) and ,
ii) and ,
iii) Make a conjecture regarding your observation in (i) and (ii),
(1 mark)
iv) By clearly showing your working evaluate . 1 mark)
(2 marks)

Ask by Bright Schneider. in Kenya

Jan 20,2025

Question

QUESTION TWO (15 MARKS)
(a) Consider a rock that is launched from the ground at a speed of . Neglecting air
resistance, a well known formula from physics states that the position of the rock after
seconds is given by the function . The position is measured in feet with
corresponding to the ground. Find the average velocity of the rock between each
pair of intervals given as
i) and ,
ii) and ,
iii) Make a conjecture regarding your observation in (i) and (ii),
(1 mark)
iv) By clearly showing your working evaluate . 1 mark)
(2 marks)

Ask by Bright Schneider. in Kenya

Jan 20,2025

UpStudy AI · Accepted Answer

**Question Two (15 MARKS)**

**(a)**

The position of the rock is given by $$ S(t) = -16t^2 + 96t $$ (in feet). To find the average velocity over different intervals, we use the formula:

$$
\text{Average Velocity} = \frac{S(t_2) - S(t_1)}{t_2 - t_1}
$$

---

**i) Between $$ t = 1 $$ sec and $$ t = 3 $$ sec**

1. **Calculate $$ S(3) $$:**
   $$
   S(3) = -16(3)^2 + 96(3) = -16(9) + 288 = -144 + 288 = 144 \text{ ft}
   $$

2. **Calculate $$ S(1) $$:**
   $$
   S(1) = -16(1)^2 + 96(1) = -16 + 96 = 80 \text{ ft}
   $$

3. **Average Velocity:**
   $$
   \frac{S(3) - S(1)}{3 - 1} = \frac{144 - 80}{2} = \frac{64}{2} = 32 \text{ ft/sec}
   $$

---

**ii) Between $$ t = 1 $$ sec and $$ t = 1.00001 $$ sec**

1. **Calculate $$ S(1.00001) $$:**
   $$
   S(1.00001) = -16(1.00001)^2 + 96(1.00001)
   $$
   Approximating $$ (1.00001)^2 \approx 1.00002 $$:
   $$
   S(1.00001) \approx -16(1.00002) + 96(1.00001) = -16.00032 + 96.00096 = 80.00064 \text{ ft}
   $$

2. **Average Velocity:**
   $$
   \frac{S(1.00001) - S(1)}{1.00001 - 1} = \frac{80.00064 - 80}{0.00001} = \frac{0.00064}{0.00001} = 64 \text{ ft/sec}
   $$

---

**iii) Conjecture Regarding Observations in (i) and (ii)**

From part (i), the average velocity over a larger interval (2 seconds) is **32 ft/sec**, while from part (ii), the average velocity over an infinitesimally small interval approaches **64 ft/sec**. This suggests that as the time interval decreases, the average velocity approaches a specific value, indicating the **instantaneous velocity** at $$ t = 1 $$ sec.

---

**iv) Evaluating $$ \lim_{t \to 1} \frac{S(t) - S(1)}{t - 1} $$**

This limit represents the derivative of $$ S(t) $$ at $$ t = 1 $$, which gives the instantaneous velocity.

1. **Express the Difference Quotient:**
   $$
   \frac{S(t) - S(1)}{t - 1} = \frac{-16t^2 + 96t - (-16(1)^2 + 96(1))}{t - 1} = \frac{-16t^2 + 96t - 80}{t - 1}
   $$

2. **Factor the Numerator:**
   $$
   -16t^2 + 96t - 80 = -16(t^2 - 6t + 5) = -16(t - 5)(t - 1)
   $$

3. **Simplify the Expression:**
   $$
   \frac{-16(t - 5)(t - 1)}{t - 1} = -16(t - 5) \quad \text{for } t \neq 1
   $$

4. **Take the Limit as $$ t \to 1 $$:**
   $$
   \lim_{t \to 1} -16(t - 5) = -16(1 - 5) = -16(-4) = 64 \text{ ft/sec}
   $$

---

**Summary of Results:**
- **i)** Average velocity between 1 sec and 3 sec: **32 ft/sec**
- **ii)** Average velocity between 1 sec and 1.00001 sec: **64 ft/sec**
- **iii)** The average velocity approaches **64 ft/sec** as the interval around $$ t = 1 $$ sec becomes smaller, indicating the **instantaneous velocity** at that moment.
- **iv)** $$ \lim_{t \to 1} \frac{S(t) - S(1)}{t - 1} = 64 \text{ ft/sec} $$
**Question Two (15 MARKS)** **(a)** The position of the rock is given by $$ S(t) = -16t^2 + 96t $$ (in feet). To find the average velocity over different intervals, use the formula: $$ \text{Average Velocity} = \frac{S(t_2) - S(t_1)}{t_2 - t_1} $$ --- **i) Between $$ t = 1 $$ sec and $$ t = 3 $$ sec** - **Calculate $$ S(3) $$:** $$ S(3) = -16(3)^2 + 96(3) = -144 + 288 = 144 \text{ ft} $$ - **Calculate $$ S(1) $$:** $$ S(1) = -16(1)^2 + 96(1) = -16 + 96 = 80 \text{ ft} $$ - **Average Velocity:** $$ \frac{144 - 80}{3 - 1} = \frac{64}{2} = 32 \text{ ft/sec} $$ --- **ii) Between $$ t = 1 $$ sec and $$ t = 1.00001 $$ sec** - **Calculate $$ S(1.00001) $$:** $$ S(1.00001) \approx -16(1.00001)^2 + 96(1.00001) \approx -16.00032 + 96.00096 = 80.00064 \text{ ft} $$ - **Average Velocity:** $$ \frac{80.00064 - 80}{0.00001} = \frac{0.00064}{0.00001} = 64 \text{ ft/sec} $$ --- **iii) Conjecture** As the time interval decreases, the average velocity approaches **64 ft/sec**, indicating the **instantaneous velocity** at $$ t = 1 $$ sec. --- **iv) Evaluating $$ \lim_{t \to 1} \frac{S(t) - S(1)}{t - 1} $$** This limit represents the derivative of $$ S(t) $$ at $$ t = 1 $$, giving the instantaneous velocity. - **Difference Quotient:** $$ \frac{S(t) - S(1)}{t - 1} = \frac{-16t^2 + 96t - 80}{t - 1} = -16(t - 5) $$ - **Take the Limit:** $$ \lim_{t \to 1} -16(t - 5) = -16(-4) = 64 \text{ ft/sec} $$ --- **Summary:** - **i)** Average velocity between 1 sec and 3 sec: **32 ft/sec** - **ii)** Average velocity between 1 sec and 1.00001 sec: **64 ft/sec** - **iii)** The average velocity approaches **64 ft/sec** as the interval around $$ t = 1 $$ sec becomes smaller. - **iv)** $$ \lim_{t \to 1} \frac{S(t) - S(1)}{t - 1} = 64 \text{ ft/sec} $$

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