A particle moves along a horizontal line so that at any time t its position is given by x(t)=cost-t. Time is measured in seconds and x is measured in meters.

a.) Find the velocity as a function t. Use your answer to determine the velocity of the particle when t=pi/6 seconds. Indicate units of measure.
b.) Find the acceleration as a function of the time t. Use your answer to determine the velocity of the particle when t=pi/6 seconds. Indicate units of measure.
c.) What are the values of t,0≤t≤2pi, for which the particle is at rest?
d.) What are the values of t, 0≤t≤2pi, for which the particle moves to the right?
e.) At t=pi/6 seconds, is the particle speeding up or slowing down? Justify your answer.

x = cos t - t

dx/dt = v = -sin t - 1
sin pi/6 = sin 30 = 1/2
so
v = -1.5

a = dv/dt = -cos t
I think you want the acceleration a = -sqrt3/2

when is v = 0 ?
when sin t = -1 which is t = 3 pi/2

when is v >0? when sin t <-1 which is never

when t = pi/6 , a is negative so v is getting less negative so slowing down

To solve this problem, we need to recall the definitions of velocity and acceleration.

a.) Velocity is the derivative of position with respect to time. To find the velocity as a function of t, we take the derivative of the position function x(t):
v(t) = d/dt (x(t)) = d/dt (cos(t) - t)

To determine the velocity of the particle when t = π/6 seconds, we substitute π/6 into the velocity function:
v(π/6) = d/dt (cos(π/6) - π/6)

b.) Acceleration is the derivative of velocity with respect to time. To find the acceleration as a function of t, we take the derivative of the velocity function obtained in part a:
a(t) = d/dt (v(t)) = d/dt (d/dt (cos(t) - t))

To determine the acceleration of the particle when t = π/6 seconds, we substitute π/6 into the acceleration function:
a(π/6) = d/dt (d/dt (cos(π/6) - π/6))

c.) The particle is at rest when the velocity is zero. To find the values of t for which the particle is at rest, we set the velocity function equal to zero and solve for t:
v(t) = 0

d.) The particle moves to the right when the velocity is positive. To find the values of t for which the particle moves to the right, we find the interval(s) where the velocity function is positive:
v(t) > 0

e.) To determine if the particle is speeding up or slowing down at t = π/6 seconds, we look at the sign of the acceleration at that time:
a(π/6) > 0 indicates that the particle is speeding up
a(π/6) < 0 indicates that the particle is slowing down

Now, let's plug in the values and calculate the answers step by step.

To find the velocity, we need to take the derivative of the position function x(t). Let's do this step by step.

a.) Finding the velocity as a function of t:
Given: x(t) = cost - t
Velocity is the derivative of x(t) with respect to t:
v(t) = d/dt (cost - t)
To find this derivative, we use the differentiation rules:

v(t) = -sin(t) - 1

The velocity function is v(t) = -sin(t) - 1. The units of velocity are meters per second.

Now we can determine the velocity of the particle when t = π/6 seconds by substituting π/6 into the velocity function:

v(π/6) = -sin(π/6) - 1
= -1/2 - 1
= -3/2

Therefore, when t = π/6 seconds, the velocity of the particle is -3/2 meters per second.

b.) Finding the acceleration as a function of t:
Acceleration is the derivative of velocity with respect to t. Let's differentiate the velocity function v(t):

a(t) = d/dt (-sin(t) - 1)
= -cos(t)

The acceleration function is a(t) = -cos(t). The units of acceleration are meters per second squared.

Now we can determine the velocity of the particle when t = π/6 seconds by substituting π/6 into the acceleration function:

a(π/6) = -cos(π/6)
= -√3/2

Therefore, when t = π/6 seconds, the acceleration of the particle is -√3/2 meters per second squared.

c.) To find when the particle is at rest, we need to find the values of t where the velocity is zero. Let's set v(t) = 0 and solve for t:

0 = -sin(t) - 1
1 = -sin(t)
sin(t) = -1

The values of t for which sin(t) = -1 are t = (3π/2) + n(2π), where n is an integer.

d.) To find when the particle moves to the right, we need to find the values of t where the velocity is positive. Let's set v(t) > 0 and solve for t:

-sin(t) - 1 > 0
-sin(t) > 1
sin(t) < -1

There are no values of t for which sin(t) < -1. Therefore, the particle never moves to the right.

e.) To determine if the particle is speeding up or slowing down at t = π/6 seconds, we need to examine the sign of the acceleration.

a(π/6) = -cos(π/6)
= -√3/2

Since the acceleration is negative, the particle is slowing down at t = π/6 seconds.