Question 14 (Mandatory) (1 point) Saved A bike accelerates from \( 100 \mathrm{~km} / \mathrm{hr} \) to \( 200 \mathrm{~km} / \mathrm{hr} \) in 4 hours. What is the acceleration? \( 0.020 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) \( 25 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) \( 0.002 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) \( 10 \mathrm{~m} / \mathrm{s} / \mathrm{s} \)

To find the acceleration, we can use the formula for acceleration: \[ a = \frac{\Delta v}{\Delta t} \] where: - \( \Delta v \) is the change in velocity, - \( \Delta t \) is the change in time. First, we need to convert the velocities from kilometers per hour (km/h) to meters per second (m/s): 1. Convert \( 100 \, \text{km/h} \) to m/s: \[ 100 \, \text{km/h} = \frac{100 \times 1000 \, \text{m}}{3600 \, \text{s}} \approx 27.78 \, \text{m/s} \] 2. Convert \( 200 \, \text{km/h} \) to m/s: \[ 200 \, \text{km/h} = \frac{200 \times 1000 \, \text{m}}{3600 \, \text{s}} \approx 55.56 \, \text{m/s} \] Now, calculate the change in velocity (\( \Delta v \)): \[ \Delta v = 55.56 \, \text{m/s} - 27.78 \, \text{m/s} = 27.78 \, \text{m/s} \] The time interval (\( \Delta t \)) is given as 4 hours, which we need to convert to seconds: \[ 4 \, \text{hours} = 4 \times 3600 \, \text{s} = 14400 \, \text{s} \] Now we can calculate the acceleration: \[ a = \frac{\Delta v}{\Delta t} = \frac{27.78 \, \text{m/s}}{14400 \, \text{s}} \approx 0.00193 \, \text{m/s}^2 \] This value is approximately \( 0.002 \, \text{m/s}^2 \). Thus, the correct answer is: \[ \boxed{0.002 \, \text{m/s}^2} \]