Particle Problem

[Jason76]

A particle is moving on a vertical line so that its coordinate at time t is \(\displaystyle t^{3} - 12t + 3, [0, \infty)\)

a. Find the velocity function.

b. Find the maximum height of the particle on the interval \(\displaystyle [0,3]\)

c. Find the acceleration function.

a.

The first derivative of the time function is the velocity function.

\(\displaystyle f(t) = t^{3} - 12t + 3\)

\(\displaystyle f'(t) = v(t) = 3t^{2} - 12\)

b.

Find critical numbers, plug them, and the sides of the interval \(\displaystyle [0,3]\) into the main time function to find the maximum (highest y value).

\(\displaystyle f(t) = t^{3} - 12t + 3\)

\(\displaystyle t^{3} - 12t + 3 = 0\) How do I solve this for \(\displaystyle t\)?

c.

The second derivative of the time function is the acceleration function.

\(\displaystyle f''(t) = a(t) = 6t\)

 

[stapel]

A particle is moving on a vertical line so that its coordinate at time t is t3−12t+3,[0,∞)
a. Find the velocity function.
b. Find the maximum height of the particle on the interval [0,3]
c. Find the acceleration function.

b. Find critical numbers, plug them, and the sides of the interval \(\displaystyle [0,3]\) into the main time function to find the maximum (highest y value).

\(\displaystyle f(t) = t^{3} - 12t + 3\)

\(\displaystyle t^{3} - 12t + 3 = 0\) How do I solve this for \(\displaystyle t\)?

You don't. The critical numbers of a function are not the zeroes of the original function. (Hint: Check the definition of "critical number".)

 

[Jason76]

Small mistake. The critical numbers are the result of the 1st derivative set to \(\displaystyle 0\)