8. A ball rolls down an inclined track and gains speed. If the distance the ball travels is 3 cm in the first second, 6 cm in the second, 9 cm in the third, and so on, find the total distance travelled in 15 seconds. ( 3 marks) 9. A superball bounces to $$ \frac{3}{4} $$ of its initial height when dropped on pavement. If the ball is dropped from a height of 16 m , how far has the ball travelled by the!time it hits the (4 marks) (4) for the $$ 6^{ ext {th }} $$ time?

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8. A ball rolls down an inclined track and gains speed. If the distance the ball travels is 3 cm in the first second, 6 cm in the second, 9 cm in the third, and so on, find the total distance travelled in 15 seconds. ( 3 marks) 9. A superball bounces to $$ \frac{3}{4} $$ of its initial height when dropped on pavement. If the ball is dropped from a height of 16 m , how far has the ball travelled by the!time it hits the (4 marks) (4) for the $$ 6^{	ext {th }} $$ time?

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Sure, let's tackle each problem step by step.

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### **Question 8**

**A ball rolls down an inclined track and gains speed. If the distance the ball travels is 3 cm in the first second, 6 cm in the second, 9 cm in the third, and so on, find the total distance travelled in 15 seconds. (3 marks)**

**Solution:**

We observe that the distance traveled each second forms an arithmetic sequence where:
- **First term (a₁)** = 3 cm
- **Common difference (d)** = 3 cm (since each second the distance increases by 3 cm)

We need to find the **total distance (Sₙ)** traveled in **n = 15 seconds**.

The formula for the sum of the first n terms of an arithmetic sequence is:
$$ S_n = \frac{n}{2} \times [2a_1 + (n - 1)d] $$

Plugging in the values:
$$ S_{15} = \frac{15}{2} \times [2 \times 3 + (15 - 1) \times 3] $$
$$ S_{15} = \frac{15}{2} \times [6 + 42] $$
$$ S_{15} = \frac{15}{2} \times 48 $$
$$ S_{15} = 15 \times 24 $$
$$ S_{15} = 360 \text{ cm} $$

**Answer:** The total distance traveled in 15 seconds is **360 cm**.

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### **Question 9**

**A superball bounces to $$ \frac{3}{4} $$ of its initial height when dropped on pavement. If the ball is dropped from a height of 16 m, how far has the ball traveled by the time it hits the 6\textsuperscript{th} time? (4 marks)**

**Solution:**

Let's analyze the ball's motion:

1. **Initial Drop:**
   - Falls 16 m to the pavement. (First hit)

2. **Subsequent Bounces:**
   - After each bounce, it rises to $$ \frac{3}{4} $$ of the previous height.
   - Then it falls the same distance back to the pavement.

We need to calculate the total distance up to the 6\textsuperscript{th} hit.

**Breakdown of Bounces:**

- **Hit 1:** Drops 16 m.
- **Bounce 1:** Rises to $$ 16 \times \frac{3}{4} = 12 $$ m, then falls 12 m. (Hits 2)
- **Bounce 2:** Rises to $$ 12 \times \frac{3}{4} = 9 $$ m, then falls 9 m. (Hits 3)
- **Bounce 3:** Rises to $$ 9 \times \frac{3}{4} = 6.75 $$ m, then falls 6.75 m. (Hits 4)
- **Bounce 4:** Rises to $$ 6.75 \times \frac{3}{4} = 5.0625 $$ m, then falls 5.0625 m. (Hits 5)
- **Bounce 5:** Rises to $$ 5.0625 \times \frac{3}{4} = 3.796875 $$ m, then falls 3.796875 m. (Hits 6)

**Total Distance Calculation:**

- **Initial Drop:** 16 m
- **Each Bounce (Up and Down):**
  - Bounce 1: 12 m + 12 m = 24 m
  - Bounce 2: 9 m + 9 m = 18 m
  - Bounce 3: 6.75 m + 6.75 m = 13.5 m
  - Bounce 4: 5.0625 m + 5.0625 m = 10.125 m
  - Bounce 5: 3.796875 m + 3.796875 m = 7.59375 m

**Summing All Distances:**
$$ \text{Total Distance} = 16 + 24 + 18 + 13.5 + 10.125 + 7.59375 $$
$$ \text{Total Distance} = 16 + 24 = 40 $$
$$ 40 + 18 = 58 $$
$$ 58 + 13.5 = 71.5 $$
$$ 71.5 + 10.125 = 81.625 $$
$$ 81.625 + 7.59375 = 89.21875 \text{ m} $$

**Rounded to Three Decimal Places:**
$$ \text{Total Distance} \approx 89.219 \text{ m} $$

**Answer:** By the time it hits the 6\textsuperscript{th} time, the ball has traveled approximately **89.219 meters**.

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**Question 8:** The ball travels a total of **360 cm** in 15 seconds. **Question 9:** By the 6th hit, the ball has traveled approximately **89.219 meters**.

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