Math Problem Statement

Suppose the position of an object moving horizontally after t seconds is given by the following function sequals​f(t), where s is measured in​ feet, with sgreater than0 corresponding to positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object​ stationary, moving to the​ right, and moving to the​ left? c. Determine the velocity and acceleration of the object at tequals1. d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed​ increasing? ​f(t)equalst cubed minus 12 t squared plus 45 t​; 0less than or equalstless than or equals7

Solution

Here are the graphs for the position, velocity, and acceleration functions:

  • Position Function: The graph of f(t)=t312t2+45tf(t) = t^3 - 12t^2 + 45t shows how the object's position changes over time.
  • Velocity Function: The graph of v(t)=3t224t+45v(t) = 3t^2 - 24t + 45 represents the object's velocity, indicating when the object is stationary, moving to the right, or moving to the left.
  • Acceleration Function: The graph of a(t)=6t24a(t) = 6t - 24 shows the object's acceleration over time.

Now, let's continue with solving for the stationary points and analyzing when the object is moving right, left, and stationary.

Finding stationary points (where v(t)=0v(t) = 0):

3t224t+45=03t^2 - 24t + 45 = 0

We'll solve this equation and proceed with the rest of the calculations.The object is stationary at t=3t = 3 seconds and t=5t = 5 seconds. These are the points where the velocity is zero, meaning the object temporarily stops moving.

  • For 0t<30 \leq t < 3, the velocity is positive (object moves to the right).
  • For 3<t<53 < t < 5, the velocity is negative (object moves to the left).
  • For 5<t75 < t \leq 7, the velocity is positive again (object moves to the right).

Next, I'll calculate the velocity and acceleration at t=1t = 1, as well as the acceleration when the velocity is zero.Here are the results for the remaining parts:

c. Velocity and Acceleration at t=1t = 1:

  • Velocity at t=1t = 1: v(1)=24v(1) = 24 feet per second.
  • Acceleration at t=1t = 1: a(1)=18a(1) = -18 feet per second squared.

d. Acceleration When Velocity is Zero:

  • At t=3t = 3 seconds (first stationary point): a(3)=6a(3) = -6 feet per second squared.
  • At t=5t = 5 seconds (second stationary point): a(5)=6a(5) = 6 feet per second squared.

e. Intervals Where Speed is Increasing:

The speed increases when velocity and acceleration have the same sign:

  • From t=0t = 0 to t=3t = 3, both velocity and acceleration are positive, so the speed is increasing.
  • From t=5t = 5 to t=7t = 7, both velocity and acceleration are positive again, so the speed is increasing.

Would you like more details or have any further questions?


Here are five related questions for further exploration:

  1. How would you find the total distance traveled by the object over the interval 0t70 \leq t \leq 7?
  2. How does the acceleration function influence the shape of the velocity curve?
  3. What physical scenarios could cause an object to have positive velocity but negative acceleration?
  4. Can you determine when the object's motion is at its fastest?
  5. How would friction affect the motion described by the given equations?

Tip: The second derivative (acceleration) often reveals key inflection points, indicating changes in the nature of the movement.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Kinematics

Formulas

Position function: s(t) = t^3 - 12t^2 + 45t
Velocity function: v(t) = 3t^2 - 24t + 45
Acceleration function: a(t) = 6t - 24

Theorems

First derivative as velocity
Second derivative as acceleration

Suitable Grade Level

Grades 11-12, Early College