A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by

Question

A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by

𝑣(𝑡) = (2𝜋− 5)𝑡− sin(𝜋𝑡)
A. Find the acceleration at any time t.
B. Find the minimum acceleration of the particle over the interval [0, 3].
C. Find the maximum velocity of the particle over the interval [0, 2].

Answer 1

A. no difficulty here.

a(t) = v'(t) = (2π-5) - πcos(πt)
B. Now find where a'(t) = 0 and a"(t) > 0
C. find where v'=0 and v" < 0

You can check your work by looking at the graph at

https://www.wolframalpha.com/input/?i=plot+%282%CF%80-5%29t+-+sin%28%CF%80t%29%2C+%282%CF%80-5%29+-+%CF%80*cos%28%CF%80t%29+for+0+%3C%3D+t+%3C%3D+3

Answer 2

To find the acceleration at any time t, we can differentiate the velocity function to get the acceleration function. The acceleration is the derivative of the velocity with respect to time.

A. Find the acceleration at any time t:
We have the velocity function 𝑣(𝑡) = (2𝜋− 5)𝑡− sin(𝜋𝑡).
To find the acceleration, differentiate 𝑣(𝑡) with respect to t using the product rule and the chain rule:

𝑎(𝑡) = 𝑣'(𝑡)
= [(2𝜋− 5)𝑡]' - [sin(𝜋𝑡)]'

Differentiating each term separately, we get:

𝑎(𝑡) = [(2𝜋− 5)𝑡]' - [sin(𝜋𝑡)]'
= (2𝜋 - 5)'𝑡 + (2𝜋 - 5)(𝑡)' - sin'(𝜋𝑡)
= 0 + (2𝜋 - 5) - 𝜋cos(𝜋𝑡)

So, the acceleration at any time t is given by 𝑎(𝑡) = (2𝜋 - 5) - 𝜋cos(𝜋𝑡).

B. Find the minimum acceleration of the particle over the interval [0, 3]:
To find the minimum acceleration over the interval [0, 3], we need to determine the critical points of the acceleration function 𝑎(𝑡) = (2𝜋 - 5) - 𝜋cos(𝜋𝑡) and then evaluate them to find the minimum.

First, let's find the critical points by setting the derivative of 𝑎(𝑡) equal to zero:

𝑎'(𝑡) = -𝜋sin(𝜋𝑡) = 0

Setting -𝜋sin(𝜋𝑡) = 0, we find that 𝑡 = 0 and 𝑡 = 1 are critical points.

Next, we need to evaluate 𝑎(𝑡) at the critical points and the endpoints of the interval [0, 3]:

At 𝑡 = 0: 𝑎(0) = (2𝜋 - 5) - 𝜋cos(𝜋(0)) = (2𝜋 - 5) - 𝜋cos(0) = 2𝜋 - 5 - 𝜋

At 𝑡 = 1: 𝑎(1) = (2𝜋 - 5) - 𝜋cos(𝜋(1)) = (2𝜋 - 5) - 𝜋cos(𝜋) = 2𝜋 - 5 + 𝜋

At 𝑡 = 3: 𝑎(3) = (2𝜋 - 5) - 𝜋cos(𝜋(3)) = (2𝜋 - 5) - 𝜋cos(3𝜋) = 2𝜋 - 5 - 𝜋

Evaluating these values, we see that 𝑎(0) = 𝑎(3) = 2𝜋 - 5 - 𝜋 and 𝑎(1) = 2𝜋 - 5 + 𝜋.

The minimum acceleration over the interval [0, 3] occurs at 𝑡 = 0 and 𝑡 = 3, where 𝑎(0) = 𝑎(3) = 2𝜋 - 5 - 𝜋.