The velocity of a particle moving along the x-axis at any time t ≥ 0 is given by

a. find the particles maximum acceleration over the interval [0,2]. use derivatives to support and justify your answer.

b. find the particles minimum velocity over the interval [0,3] justify your answer.

i know that acceleration is the derivative of velocity and that:

but i am confused as to actually figure out the minimum and maximum values using cos and sin

A. a(3/2) = 3π − 7

B. v(3) = 6π − 22

Step-by-step explanation:

As you found:

v(t) = cos(πt) − t (7 − 2π)

v'(t) = a(t) = -π sin(πt) − 7 + 2π

v"(t) = a'(t) = -π² cos(πt)

a) When the acceleration is a maximum, a'(t) = 0.

0 = -π² cos(πt)

0 = cos(πt)

πt = π/2 + 2kπ, 3π/2 + 2kπ

πt = π/2 + kπ

t = 1/2 + k

t = 1/2, 3/2

We need to check if these are minimums or maximums.  To do that, we evaluate the sign of a'(t) within the intervals before and after each value.

a'(0) = -π² cos(0) = -π²

a'(1) = -π² cos(π) = π²

a'(2) = -π² cos(2π) = -π²

At t = 1/2, a'(t) changes signs from - to +.  At t = 3/2, a'(t) changes signs from + to -.  Therefore, t = 1/2 is a local minimum and t = 3/2 is a local maximum.  At t = 3/2, the acceleration is:

a(3/2) = -π sin(3π/2) − 7 + 2π

a(3/2) = 3π − 7

Compare to the endpoints:

a(0) = -π sin(0) − 7 + 2π = -7 + 2π

a(2) = -π sin(2π) − 7 + 2π = -7 + 2π

So a(3/2) = 3π − 7 is the global maximum.

b) Use the same steps as before.  When the velocity is a minimum, a(t) = 0.

0 = -π sin(πt) − 7 + 2π

π sin(πt) = -7 + 2π

sin(πt) = -7/π + 2

πt ≈ 3.372 + 2kπ, 6.053 + 2kπ

t ≈ 1.073 + 2k, 1.927 + 2k

t ≈ 1.073, 1.927

Now we evaluate the sign of a(t) in the intervals before and after each value.

a(0) = -π sin(0) − 7 + 2π = -7 + 2π

a(3/2) = -π sin(3π/2) − 7 + 2π = -7 + 3π

a(2) = -π sin(2π) − 7 + 2π = -7 + 2π

At t = 1.073, a(t) changes signs from - to +.  At t = 1.927, a(t) changes signs from + to -.  Therefore. t = 1.073 is a local minimum and t = 1.927 is a local maximum.  At t = 1.073, the velocity is:

v(1.073) = cos(1.073π) − 1.073 (7 − 2π)

v(1.073) = -1.743

Compare to the endpoints:

v(0) = cos(0) − 0 (7 − 2π) = 1

v(3) = cos(3π) − 3 (7 − 2π) = -22 + 6π

Here, v(3) = -22 + 6π is the global minimum.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

answer: 0,6

step-by-step explanation:

this would make it the y intercept.