A particle is moving at some constant speed. Use the following information to answer the following questions.

s(t)=100/t©÷+12; t¡Ã0; see represents feet and t represents seconds
a)Find the velocity and acceleration equations for this particle.
b)When is this particle at rest?
c) On what interval(s) is this particle speeding up?

I cannot interpret your symbols

100/t©÷+12; t¡Ã0

Can you think of a better way to type that equation?

a) To find the velocity, we need to find the derivative of the position equation.

s(t) = 100/t + 12

Taking the derivative with respect to time, we get:

v(t) = d/dt (100/t + 12)

Using the power rule of differentiation and chain rule, the derivative of 100/t is -100/t^2. The derivative of 12 is 0 since it is a constant. Therefore, the velocity equation is:

v(t) = -100/t^2

To find the acceleration, we need to find the derivative of the velocity equation.

a(t) = d/dt (-100/t^2)

Using the power rule of differentiation and chain rule, the derivative of -100/t^2 is 200/t^3. Therefore, the acceleration equation is:

a(t) = 200/t^3

b) The particle is at rest when the velocity is zero. So, we need to find when v(t) = 0.

-100/t^2 = 0

Since the constant -100 cannot be zero, the equation is only satisfied when t is infinity (∞). Therefore, the particle is at rest at t = ∞.

c) The particle is speeding up when the acceleration is greater than zero. So, we need to find when a(t) > 0.

200/t^3 > 0

Since the constant 200 cannot be zero, and t cannot be zero either, the inequality is only satisfied when t > 0. Therefore, the particle is speeding up on the interval (0, ∞).

To find the velocity and acceleration equations for the particle, we need to take the derivative of the equation for position, s(t), with respect to time, t.

a) Velocity equation:
The velocity of the particle is the derivative of the position equation, s(t), with respect to time, t. Taking the derivative:

s(t) = 100/t + 12

To find the velocity, we take the derivative of s(t) with respect to t:

v(t) = d/dt (100/t + 12) = -100/t^2

So, the velocity equation for this particle is v(t) = -100/t^2.

b) Finding when the particle is at rest:
A particle is at rest when its velocity is zero. So, we set the velocity equation v(t) = -100/t^2 equal to zero and solve for t:

-100/t^2 = 0

Since the numerator is zero, the equation is equivalent to 0 = 0, which is true for all values of t. This means that the particle is always at rest.

c) Determining when the particle is speeding up:
To determine when the particle is speeding up, we need to analyze the acceleration. The acceleration is the derivative of the velocity equation, v(t), with respect to time, t.

Acceleration equation:
a(t) = d/dt (-100/t^2) = 200/t^3.

To find when the particle is speeding up, we need to consider the signs of the acceleration equation. In this case, the acceleration equation a(t) = 200/t^3 is always positive since both the numerator and denominator are positive. Therefore, the particle is speeding up on the entire interval t > 0.

In summary:
a) The velocity equation for this particle is v(t) = -100/t^2.
b) The particle is always at rest.
c) The particle is speeding up on the interval t > 0.