classroom Dashboard DeltaMath Student... Pear Assessment Physics Classroom IM3 Spring 2025 - G... A car coasts into a hill at $$ 16.0 \mathrm{~m} / \mathrm{s} $$. It slows down with a uniform acceleration of $$ -1.0 \mathrm{~m} / \mathrm{s} / \mathrm{s} $$. a. What is the car's speed after 7.0 s ? Speed 9 $$ \mathrm{m} / \mathrm{s} $$ b. What is the car's displacement after 7.0 s? $$ \square $$ Displacement 87.5 m Info Attempts: 2/00 c. What is the car's speed after 10.0 s? Speed 6 $$ \mathrm{m} / \mathrm{s} $$ d. What is the car's displacement after 10.0 s? Displacement m Info Attempts: 3/00 -35 1.55 0.64

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**Problem Statement:**
A car is coasting up a hill with an initial speed of $$ 16.0 \, \text{m/s} $$. It slows down with a uniform (constant) acceleration of $$ -1.0 \, \text{m/s}^2 $$.

---

### a. What is the car's speed after 7.0 seconds?

**Given:**
- Initial speed ($$ u $$) = $$ 16.0 \, \text{m/s} $$
- Acceleration ($$ a $$) = $$ -1.0 \, \text{m/s}^2 $$
- Time ($$ t $$) = $$ 7.0 \, \text{s} $$

**Formula:**
The final speed ($$ v $$) after time $$ t $$ can be found using the equation:
$$
v = u + a \cdot t
$$

**Calculation:**
$$
v = 16.0 \, \text{m/s} + (-1.0 \, \text{m/s}^2) \times 7.0 \, \text{s} = 16.0 - 7.0 = 9.0 \, \text{m/s}
$$

**Answer:**
After 7.0 seconds, the car's speed is **9.0 m/s**.

---

### b. What is the car's displacement after 7.0 seconds?

**Given:**
- Initial speed ($$ u $$) = $$ 16.0 \, \text{m/s} $$
- Acceleration ($$ a $$) = $$ -1.0 \, \text{m/s}^2 $$
- Time ($$ t $$) = $$ 7.0 \, \text{s} $$

**Formula:**
Displacement ($$ s $$) can be calculated using:
$$
s = u \cdot t + \frac{1}{2} a t^2
$$

**Calculation:**
$$
s = 16.0 \, \text{m/s} \times 7.0 \, \text{s} + \frac{1}{2} \times (-1.0 \, \text{m/s}^2) \times (7.0 \, \text{s})^2
$$
$$
s = 112.0 \, \text{m} + \frac{1}{2} \times (-1.0) \times 49
$$
$$
s = 112.0 \, \text{m} - 24.5 \, \text{m} = 87.5 \, \text{m}
$$

**Answer:**
After 7.0 seconds, the car's displacement is **87.5 meters**.

---

### c. What is the car's speed after 10.0 seconds?

**Given:**
- Initial speed ($$ u $$) = $$ 16.0 \, \text{m/s} $$
- Acceleration ($$ a $$) = $$ -1.0 \, \text{m/s}^2 $$
- Time ($$ t $$) = $$ 10.0 \, \text{s} $$

**Formula:**
$$
v = u + a \cdot t
$$

**Calculation:**
$$
v = 16.0 \, \text{m/s} + (-1.0 \, \text{m/s}^2) \times 10.0 \, \text{s} = 16.0 - 10.0 = 6.0 \, \text{m/s}
$$

**Answer:**
After 10.0 seconds, the car's speed is **6.0 m/s**.

---

### d. What is the car's displacement after 10.0 seconds?

**Given:**
- Initial speed ($$ u $$) = $$ 16.0 \, \text{m/s} $$
- Acceleration ($$ a $$) = $$ -1.0 \, \text{m/s}^2 $$
- Time ($$ t $$) = $$ 10.0 \, \text{s} $$

**Formula:**
$$
s = u \cdot t + \frac{1}{2} a t^2
$$

**Calculation:**
$$
s = 16.0 \, \text{m/s} \times 10.0 \, \text{s} + \frac{1}{2} \times (-1.0 \, \text{m/s}^2) \times (10.0 \, \text{s})^2
$$
$$
s = 160.0 \, \text{m} + \frac{1}{2} \times (-1.0) \times 100
$$
$$
s = 160.0 \, \text{m} - 50.0 \, \text{m} = 110.0 \, \text{m}
$$

**Answer:**
After 10.0 seconds, the car's displacement is **110.0 meters**.

---

**Summary of Answers:**
- **a.** 9.0 m/s
- **b.** 87.5 m
- **c.** 6.0 m/s
- **d.** 110.0 m

If you have any further questions or need additional clarification on any of these steps, feel free to ask!
After 7.0 seconds, the car's speed is 9.0 m/s and its displacement is 87.5 meters. After 10.0 seconds, the car's speed is 6.0 m/s and its displacement is 110.0 meters.

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