Question

b) the position of a bug moving along a straight path is given by s(t)=t² - 2t + 3. at what time t is the instantaneous velocity equal to the average velocity of the bug on 0,6?

Explanation:

Step1: Find the average velocity

The formula for average velocity \(v_{avg}=\frac{s(b)-s(a)}{b - a}\). Here \(a = 0\), \(b=6\), and \(s(t)=t^{2}-2t + 3\).
First, calculate \(s(6)\):
\(s(6)=6^{2}-2\times6 + 3=36-12 + 3=27\)
Then calculate \(s(0)\):
\(s(0)=0^{2}-2\times0 + 3=3\)
So, \(v_{avg}=\frac{s(6)-s(0)}{6-0}=\frac{27 - 3}{6}=\frac{24}{6}=4\)

Step2: Find the instantaneous velocity

The formula for instantaneous velocity is \(v(t)=s^\prime(t)\).
Differentiate \(s(t)=t^{2}-2t + 3\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(s^\prime(t)=\frac{d}{dt}(t^{2}-2t + 3)=2t-2\)

Step3: Set instantaneous velocity equal to average velocity

Set \(v(t)=v_{avg}\), so \(2t-2 = 4\)
Add \(2\) to both sides: \(2t=4 + 2=6\)
Divide both sides by \(2\): \(t = 3\)

Answer:

\(t = 3\)