Question

A particle moves along the x-axis with velocity
v(t) = -(t-3)² + 5 for [0,6].

a) Find the average velocity of this particle during the interval [0,6].

b) Find a time t* ∈ [0,6] such that the velocity at time t* is equal to the average velocity during the interval [0,6]. Is it clear that such a point exists? Is there more than one such point in this case? Use the graph of v(t) as a function of t to explain how you would find t* graphically.

Thank you so much!! I really appreciate it!

Answer 1

integrate to find x
x = -(1/3)(t-3)^3 + 5 t + c

x(6) = do calculation
x(0) = do calculation
------------------------ subtract

v av = [ x(6) - x(0) ] / 6

solve for t when v = v av

mean value theorem says at least one such point.

Answer 2

try graphing software with:

-(1/3) * (x-3)^3 + 5* x

x range -5 to + 10
y range -40 to + 40

Answer 3

for example

http://rechneronline.de/function-graphs/