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A tug is moxing hack and forth on a straight path. The velocity of the bug is given by $\newline$ $v(t)=t^{2}-3t$ . Find the average acceleration of the bug on the interval $\newline$ $[1,4]$ .

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A tug is moxing hack and forth on a straight path. The velocity of the bug is given by

\newline

v(t)=t^{2}-3t

. Find the average acceleration of the bug on the interval

\newline

[1,4]

.

Calculate Velocities: First, we need to find the change in velocity over the interval $[1,4]$ . So we calculate the velocity at $t=4$ and $t=1$ . $v(4) = 4^2 - 3\times4 = 16 - 12 = 4$ $v(1) = 1^2 - 3\times1 = 1 - 3 = -2$
Find Change in Velocity: Now, we subtract the initial velocity from the final velocity to get the change in velocity ( $\Delta v$ ). $\newline$ $\Delta v = v(4) - v(1) = 4 - (-2) = 6$
Calculate Change in Time: Next, we find the change in time ( $\Delta t$ ) over the interval $[1,4]$ . That's simply $4 - 1 = 3$ seconds.
Calculate Average Acceleration: To find the average acceleration ( $a_{\text{avg}}$ ), we divide the change in velocity ( $\Delta v$ ) by the change in time ( $\Delta t$ ). $\newline$ $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{6}{3}$

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