Question

At time t in seconds, a particle's distance s(t), in centimeters, from a point is given by
s(t) = 4 + 3sin(t)
What is the average velocity of the particle from t = π/3 to t = 7π/3?

Answer 1

The average velocity of a position function is the same as "the average rate of change of the position function." Over an interval \([a,b]\), the average rate of change of a function \(s(t)\) is given by
\[\frac{s(b)-s(a)}{b-a}\]

Answer 2

So in this case, the avg rate of change is
\[\frac{(4+\sin3\left(\frac{7\pi}{3}\right))-(4+\sin3\left(\frac{\pi}{3}\right))}{\frac{7\pi}{3}-\frac{\pi}{3}}\]

Answer 3

thanks! can you help me simplify that?

Answer 4

I'll do the denominator

its

\[2 \pi\]

Answer 5

\(\sin3\left(\dfrac{7\pi}{3}\right)=\sin7\pi=0\). Do the same with the other trig term.

Answer 6

thank you!

Answer 7

yw