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?

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!