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!
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.
try graphing software with:
-(1/3) * (x-3)^3 + 5* x
x range -5 to + 10
y range -40 to + 40
for example
http://rechneronline.de/function-graphs/
To find the average velocity of the particle during the interval [0,6], we need to calculate the displacement of the particle during that time and divide it by the time taken.
a) Average velocity is given by:
Average velocity = (Displacement) / (Time taken)
The displacement of the particle from time t=0 to t=6 is given by:
Displacement = Position at t=6 - Position at t=0
To find the position of the particle at any time t, we integrate the velocity function v(t). Integrating v(t) will give us the position function x(t).
x(t) = -(1/3)(t-3)^3 + 5t + C
Now we can find the positions at t=0 and t=6:
Position at t=0:
x(0) = -(1/3)(0-3)^3 + 5(0) + C
= 9 + C
Position at t=6:
x(6) = -(1/3)(6-3)^3 + 5(6) + C
= 0 + 30 + C
= 30 + C
So, the displacement is:
Displacement = 30 + C - (9 + C)
= 21
The time taken is t=6 - t=0 = 6.
Therefore, the average velocity is:
Average velocity = Displacement / Time taken
= 21 / 6
= 3.5
b) To find a time t* such that the velocity at time t* is equal to the average velocity during the interval [0,6], we set the velocity function equal to the average velocity and solve for t.
-(t* - 3)^2 + 5 = 3.5
Simplifying the equation, we get:
(t* - 3)^2 = 1.5
Taking the square root of both sides, we find:
t* - 3 = ±√1.5
Solving for t*, we have:
t* = 3 ± √1.5
So, there are two possible values for t*. One is t* = 3 + √1.5 and the other is t* = 3 - √1.5.
To determine if such a point exists on the graph, we can examine the graph of v(t).
The graph of v(t) is a downward-opening parabola with its vertex at (3, 5) and the y-intercept at (0, 5).
To find t* graphically, we look for the point(s) on the graph where the velocity function intersects the line y=3.5. These points correspond to the time(s) when the velocity equals the average velocity.
By visually inspecting the graph, we can see that the line y=3.5 intersects the parabola at two points. One point is to the left of the vertex and the other point is to the right of the vertex. These points correspond to the two possible values of t*, t* = 3 + √1.5 and t* = 3 - √1.5.
Therefore, there are two points in this case where the velocity at time t* is equal to the average velocity during the interval [0,6].