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A bug is moving back and forth on a straight path. The velocity of the bug is given by
v(t)=t^(2)-3t. Find the average acceleration of the bug on the interval
[1,4].

|(4^(3))/(3)-24||(t^(3))/(3)-(3t^(2))/(2)|_(1)^(4)

A bug is moving back and forth on a straight path. The velocity of the bug is given by $v(t)=t^{2}-3 t$ . Find the average acceleration of the bug on the interval $[1,4]$ . $\newline$ $\left|\frac{4^{3}}{3}-24\right|\left|\frac{t^{3}}{3}-\frac{3 t^{2}}{2}\right|_{1}^{4}$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A bug is moving back and forth on a straight path. The velocity of the bug is given by

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

. Find the average acceleration of the bug on the interval

[1,4]

.

\newline

\left|\frac{4^{3}}{3}-24\right|\left|\frac{t^{3}}{3}-\frac{3 t^{2}}{2}\right|_{1}^{4}

Find acceleration function: First, find the acceleration by taking the derivative of the velocity function $v(t) = t^2 - 3t$ . $\newline$ $a(t) = \frac{dv}{dt} = 2t - 3$ .
Calculate average acceleration: Next, calculate the average acceleration over the interval $[1, 4]$ by integrating the acceleration function from $t=1$ to $t=4$ and then dividing by the interval length. $\newline$ Average acceleration = $\frac{1}{(4-1)} \times \int_{1}^{4}(2t - 3) \, dt$ .
Integrate acceleration function: Integrate the function $2t - 3$ from $1$ to $4$ . $\int(2t - 3) dt = (t^2 - 3t) |$ from $1$ to $4$ .
Plug in limits: Plug in the limits of integration. $\newline$ $(t^2 - 3t) | \text{ from } 1 \text{ to } 4 = (4^2 - 3\times4) - (1^2 - 3\times1)$ .
Calculate values: Calculate the values. $\newline$ $(16 - 12) - (1 - 3) = 4 + 2 = 6$ .
Divide by interval length: Now divide by the interval length, which is $4 - 1 = 3$ seconds. $\newline$ Average acceleration = $\frac{6}{3} = 2 \, \text{m/s}^2$ .

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