A particle travels along the $x$ -axis such that its velocity is given by $v(t)=t^{0.4} \cos (t-4)$ . What is the average acceleration of the particle on the interval $0 \leq t \leq 5$ ? You may use a calculator and round your answer to the nearest thousandth. $\newline$ Answer:

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Relate position, velocity, speed, and acceleration using derivatives

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Q. A particle travels along the

x

-axis such that its velocity is given by

v(t)=t^{0.4} \cos (t-4)

. What is the average acceleration of the particle on the interval

0 \leq t \leq 5

? You may use a calculator and round your answer to the nearest thousandth.

\newline

Answer:

Calculate Initial Velocity: To find the average acceleration over the interval from $t = 0$ to $t = 5$ , we need to calculate the change in velocity over the change in time. The formula for average acceleration ( $a_{\text{avg}}$ ) is: $\newline$ $a_{\text{avg}} = \frac{v_{\text{final}} - v_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$ $\newline$ First, we need to find the initial and final velocities using the given velocity function $v(t) = t^{0.4}\cos(t-4)$ .
Calculate Final Velocity: Calculate the initial velocity $v_{\text{initial}}$ at $t = 0$ using the velocity function: $\newline$ $v_{\text{initial}} = v(0) = 0^{0.4}\cos(0-4) = 0 \times \cos(-4) = 0$ $\newline$ Since any number raised to a positive power multiplied by zero is zero.
Calculate Average Acceleration: Calculate the final velocity $v_{\text{final}}$ at $t = 5$ using the velocity function: $\newline$ $v_{\text{final}} = v(5) = 5^{0.4}\cos(5-4) = 5^{0.4}\cos(1)$ $\newline$ Using a calculator, we find: $\newline$ $5^{0.4} \approx 2.639$ $\newline$ $\cos(1) \approx 0.540$ $\newline$ Therefore, $v_{\text{final}} \approx 2.639 \times 0.540 \approx 1.425$
Final Result: Now we have the initial and final velocities, we can calculate the average acceleration: $\newline$ $a_{\text{avg}} = \frac{v_{\text{final}} - v_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$ $\newline$ $a_{\text{avg}} = \frac{1.425 - 0}{5 - 0}$ $\newline$ $a_{\text{avg}} = \frac{1.425}{5}$ $\newline$ $a_{\text{avg}} \approx 0.285$ $\newline$ Round the answer to the nearest thousandth.
Final Result: Now we have the initial and final velocities, we can calculate the average acceleration: $\newline$ $a_{\text{avg}} = \frac{v_{\text{final}} - v_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}$ $\newline$ $a_{\text{avg}} = \frac{1.425 - 0}{5 - 0}$ $\newline$ $a_{\text{avg}} = \frac{1.425}{5}$ $\newline$ $a_{\text{avg}} \approx 0.285$ $\newline$ Round the answer to the nearest thousandth.The average acceleration of the particle on the interval from $0$ to $5$ is approximately $0.285 \, \text{m/s}^2$ when rounded to the nearest thousandth.