Question

b) 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 over the interval (1,4).

Explanation:

Step1: Recall the formula for average acceleration

The formula for average acceleration \(a_{avg}\) over the interval \([t_1,t_2]\) is \(a_{avg}=\frac{v(t_2)-v(t_1)}{t_2 - t_1}\), where \(v(t)\) is the velocity function.

Step2: Find \(v(4)\) and \(v(1)\)

Given \(v(t)=t^{2}-3t\).
For \(t = 4\): \(v(4)=4^{2}-3\times4=16 - 12=4\).
For \(t = 1\): \(v(1)=1^{2}-3\times1=1 - 3=-2\).

Step3: Calculate the average acceleration

Substitute \(t_1 = 1\), \(t_2=4\), \(v(1)=-2\) and \(v(4) = 4\) into the formula \(a_{avg}=\frac{v(t_2)-v(t_1)}{t_2 - t_1}\).
\(a_{avg}=\frac{4-(-2)}{4 - 1}=\frac{4 + 2}{3}=\frac{6}{3}=2\).

Answer:

\(2\)