12. Calculate the following: a. If a rabbit goes from $$ 10 \mathrm{~m} / \mathrm{s} $$ to $$ 35 \mathrm{~m} / \mathrm{s} $$ while accelerating at a rate of $$ 10 \mathrm{~m} / \mathrm{s}^{2} $$, how long did it take the rabbit to increase its speed? b. A cheetah accelerated by $$ 25 \mathrm{~m} / \mathrm{s}^{2} $$ over the course of 10 seconds. If the cheetah started with a velocity of $$ 50 \mathrm{~m} / \mathrm{s} $$, what was the cheetah's final velocity? c. A car is pushed off a 20 m cliff, how long would it take to hit the ground? d. A race car accelerates at $$ 3 \mathrm{~m} / \mathrm{s}^{2} $$ for 2 seconds. If it started at $$ 28 \mathrm{~m} / \mathrm{s} $$, what is its final velocity? e. If a car drives 30 m North, and then turns around and drives 5 m South, what is the car's total displacement? What is the car's total distance traveled?

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UpStudy AI · Accepted Answer

Certainly! Let's tackle each part of Problem 12 step by step.

---

### **12. Calculate the following:**

#### **a. If a rabbit goes from $$ 10 \mathrm{~m/s} $$ to $$ 35 \mathrm{~m/s} $$ while accelerating at a rate of $$ 10 \mathrm{~m/s}^2 $$, how long did it take the rabbit to increase its speed?**

**Given:**
- Initial velocity ($$ v_i $$) = $$ 10 \, \mathrm{m/s} $$
- Final velocity ($$ v_f $$) = $$ 35 \, \mathrm{m/s} $$
- Acceleration ($$ a $$) = $$ 10 \, \mathrm{m/s}^2 $$

**To Find:**
- Time taken ($$ t $$)

**Solution:**

We can use the kinematic equation that relates velocity, acceleration, and time:

$$
v_f = v_i + a \cdot t
$$

Rearranging to solve for time ($$ t $$):

$$
t = \frac{v_f - v_i}{a}
$$

Plugging in the values:

$$
t = \frac{35 \, \mathrm{m/s} - 10 \, \mathrm{m/s}}{10 \, \mathrm{m/s}^2} = \frac{25 \, \mathrm{m/s}}{10 \, \mathrm{m/s}^2} = 2.5 \, \mathrm{seconds}
$$

**Answer:**
It took the rabbit **2.5 seconds** to increase its speed from $$ 10 \, \mathrm{m/s} $$ to $$ 35 \, \mathrm{m/s} $$.

---

#### **b. A cheetah accelerated by $$ 25 \mathrm{~m/s}^2 $$ over the course of 10 seconds. If the cheetah started with a velocity of $$ 50 \mathrm{~m/s} $$, what was the cheetah's final velocity?**

**Given:**
- Initial velocity ($$ v_i $$) = $$ 50 \, \mathrm{m/s} $$
- Acceleration ($$ a $$) = $$ 25 \, \mathrm{m/s}^2 $$
- Time ($$ t $$) = $$ 10 \, \mathrm{seconds} $$

**To Find:**
- Final velocity ($$ v_f $$)

**Solution:**

Using the same kinematic equation:

$$
v_f = v_i + a \cdot t
$$

Plugging in the values:

$$
v_f = 50 \, \mathrm{m/s} + 25 \, \mathrm{m/s}^2 \times 10 \, \mathrm{s} = 50 \, \mathrm{m/s} + 250 \, \mathrm{m/s} = 300 \, \mathrm{m/s}
$$

**Answer:**
The cheetah's final velocity was **300 m/s**.

*However, it's worth noting that a cheetah cannot physically achieve such a high velocity. This suggests either the acceleration value is hypothetical or intended for illustrative purposes.*

---

#### **c. A car is pushed off a 20 m cliff, how long would it take to hit the ground?**

**Given:**
- Height of the cliff ($$ h $$) = $$ 20 \, \mathrm{m} $$
- Initial vertical velocity ($$ v_i $$) = $$ 0 \, \mathrm{m/s} $$ (assuming the car is simply pushed off horizontally)
- Acceleration due to gravity ($$ g $$) = $$ 9.81 \, \mathrm{m/s}^2 $$ (downward)

**To Find:**
- Time to hit the ground ($$ t $$)

**Solution:**

We'll use the equation for vertical displacement under constant acceleration:

$$
h = v_i \cdot t + \frac{1}{2} g t^2
$$

Since the initial vertical velocity is zero:

$$
h = \frac{1}{2} g t^2
$$

Solving for $$ t $$:

$$
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20 \, \mathrm{m}}{9.81 \, \mathrm{m/s}^2}} = \sqrt{\frac{40}{9.81}} \approx \sqrt{4.08} \approx 2.02 \, \mathrm{seconds}
$$

**Answer:**
It would take approximately **2.02 seconds** for the car to hit the ground.

---

#### **d. A race car accelerates at $$ 3 \mathrm{~m/s}^2 $$ for 2 seconds. If it started at $$ 28 \mathrm{~m/s} $$, what is its final velocity?**

**Given:**
- Initial velocity ($$ v_i $$) = $$ 28 \, \mathrm{m/s} $$
- Acceleration ($$ a $$) = $$ 3 \, \mathrm{m/s}^2 $$
- Time ($$ t $$) = $$ 2 \, \mathrm{seconds} $$

**To Find:**
- Final velocity ($$ v_f $$)

**Solution:**

Using the kinematic equation:

$$
v_f = v_i + a \cdot t
$$

Plugging in the values:

$$
v_f = 28 \, \mathrm{m/s} + 3 \, \mathrm{m/s}^2 \times 2 \, \mathrm{s} = 28 \, \mathrm{m/s} + 6 \, \mathrm{m/s} = 34 \, \mathrm{m/s}
$$

**Answer:**
The race car's final velocity is **34 m/s**.

---

#### **e. If a car drives 30 m North, and then turns around and drives 5 m South, what is the car's total displacement? What is the car's total distance traveled?**

**Given:**
- First displacement: $$ 30 \, \mathrm{m} $$ North
- Second displacement: $$ 5 \, \mathrm{m} $$ South

**To Find:**
- Total displacement
- Total distance traveled

**Solution:**

**Total Displacement:**
Displacement is a vector quantity that takes into account the direction.

- Driving North is positive: $$ +30 \, \mathrm{m} $$
- Driving South is negative: $$ -5 \, \mathrm{m} $$
  
Total displacement ($$ \Delta x $$):

$$
\Delta x = +30 \, \mathrm{m} - 5 \, \mathrm{m} = +25 \, \mathrm{m} \, \text{North}
$$

**Total Distance Traveled:**
Distance is a scalar quantity and sums the absolute values of each movement.

$$
\text{Total distance} = |30 \, \mathrm{m}| + |5 \, \mathrm{m}| = 30 \, \mathrm{m} + 5 \, \mathrm{m} = 35 \, \mathrm{m}
$$

**Answer:**
- **Total Displacement:** $$ 25 \, \mathrm{m} $$ North
- **Total Distance Traveled:** $$ 35 \, \mathrm{m} $$

---

Feel free to ask if you have any further questions or need additional clarifications!
**a.** The rabbit took **2.5 seconds** to increase its speed from $$ 10 \, \mathrm{m/s} $$ to $$ 35 \, \mathrm{m/s} $$. **b.** The cheetah's final velocity was **300 m/s**. **c.** It would take approximately **2.02 seconds** for the car to hit the ground. **d.** The race car's final velocity is **34 m/s**. **e.** - **Total Displacement:** $$ 25 \, \mathrm{m} $$ North - **Total Distance Traveled:** $$ 35 \, \mathrm{m} $$

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