Numeric 3 points A vehicle begins to accelerate after being stopped. It's distanced traveled can be expressed by the equation \( s(t)=2.54 t^{2} \) for \( 0 \leq t \leq 20 \). What is the average speed of the vehicle 20.1 seconds after it starts moving to 21 seconds? "Round to two decimal places** 104.39

To calculate the average speed of the vehicle from \( t = 20.1 \) seconds to \( t = 21 \) seconds, we first need to find the distance traveled at both of these times. Using the distance equation \( s(t) = 2.54 t^2 \): 1. At \( t = 20.1 \): \[ s(20.1) = 2.54 \times (20.1)^2 \approx 2.54 \times 404.01 \approx 1028.10 \text{ meters} \] 2. At \( t = 21 \): \[ s(21) = 2.54 \times (21)^2 \approx 2.54 \times 441 \approx 1121.94 \text{ meters} \] Now, calculate the distance traveled between these two times: \[ \Delta s = s(21) - s(20.1) \approx 1121.94 - 1028.10 \approx 93.84 \text{ meters} \] The time interval \( \Delta t \) is \( 21 - 20.1 = 0.9 \) seconds. Now, average speed is given by: \[ \text{Average speed} = \frac{\Delta s}{\Delta t} \approx \frac{93.84}{0.9} \approx 104.27 \text{ m/s} \] Rounded to two decimal places, the average speed of the vehicle from 20.1 seconds to 21 seconds is approximately **104.27 m/s**.