# Warm Up A particle moves vertically(in inches)along the x-axis according to the position equation x(t) = t4 – 18t2 + 7t – 4, where t represents seconds.

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Warm Up A particle moves vertically(in inches)along the x-axis according to the position equation x(t) = t4 – 18t2 + 7t – 4, where t represents seconds. Find: 1) The speed when t = 1 2) The velocity when t = 2 3) The average velocity on the interval [0, 2] 4) When is the particle changing direction? How do you know? 5) When is the particle moving left? Right? 6) What is the particles displacement from t = 0 to t = 2?

Particle Motion Day 2

A particle moves along the x-axis so that its velocity at any time t is given by
(a) Find the value of (b) Given the initial position of the particle is x(0) = 8. Write a function for the position of the particle at any time t. (c) Determine the displacement of the particle from t = 0 to t = 3. (d) What relationship do you notice between the definite integral and the displacement? Explain why you think this happened?

Position and Velocity is the displacement from time a to time b (in other words the change in position) x(b) = x(a) + displacement from a to b.

A particle moves along the x-axis so that its velocity at any time t, given in seconds, is given by v(t) = t3 – 5t2 – 1. The initial position of the particle is (a) How far has the particle traveled away from its starting position at t = 5 seconds? (b) Find the position of a particle after seconds.

YOUR TURN Let v(t) = 6t2 – 8t + 3 represent the velocity of a particle moving along the x-axis after t seconds. The position of the particle at t = 0 is x(0) = 6. a) How far did the particle travel away from its original position in the first seconds? b) What is the position of the particle at t = 10 seconds?

Calculator Active A particle moves along the y-axis in such a way that the velocity of the particle is given by v(t) = etsin(t) ft/min. If y(0) = 3, what is the position of the particle at t = 5 minutes?

Let’s Review… The velocity is considered positive when the distance from the CBR is increasing and negative when the distance from the CBR is decreasing. The velocity is zero when you stop, even if it is momentary. Remember that in order to change directions you had to first stop, even if you did not think you came to a complete stop. You cannot move forward and then move backward without first stopping.

What does positive velocity look like on a distance versus time graph?
What does positive velocity look like on a velocity versus time graph? What does negative velocity look like on a distance versus time graph? What does negative velocity look like on a velocity versus time graph? What does zero velocity look like on a distance versus time graph? What does zero velocity look like on a velocity versus time graph?

A particle moves along a straight line
A particle moves along a straight line. A graph of the particle’s velocity, v(t), at time t is shown for 0 < t < 7. The graph has horizontal tangents at t = 1.8, t = 3 and t = 5.1. a) At what values of t is the particle at rest? b) At what values of t does the particle change direction? c) When, 0 < t < 7, is the particle moving right? left?

[Instantaneous acceleration] v’(t) = a(t) [Average acceleration]
Acceleration is the rate of change (slope) of velocity… Units are distance/time2 [Instantaneous acceleration] v’(t) = a(t) [Average acceleration]

The graph below shows the velocity of an object over a 10 minute period.
Determine Acceleration at t = 3 sec. Average acceleration from t = 0 sec. to t = 10 sec.

A particle moves vertically(in inches)along the y-axis according to the position equation y(t) = t3 – 6t2 + 9t – 4, where t represents seconds. 1) Find acceleration when t = Be sure to include units with your answer ) Calculate the average acceleration from time t = 0 to t = 5.

Let’s summarize… P V A acceleration position velocity Instantaneous velocity = derivative of position Average velocity = slope of the secant of position Instantaneous acceleration = derivative of velocity (or second derivative of position) Average acceleration = slope of the secant line of velocity

Speeding Up or Slowing Down?
If the velocity is positive… Positive acceleration means you are speeding up and Negative acceleration means you are slowing down If the velocity is negative… Negative acceleration means you are speeding up and Positive acceleration means you are slowing down.

A particle moves along the y-axis so that it’s position at time t is given by ) Write a function for the velocity. 2) When is the particle changing direction? 3) Is the particle slowing down, speeding up, or maintaining speed at t = 1?

CALCULATOR ACTIVE A particle moves on the x-axis with velocity at time t > 0 given as v(t) = -1 + e1 - t cm/min. The position of the particle at t = 1 is -3. 1) What is the velocity of the particle at t = 3? 2) What is the average acceleration of the particle in the interval 0 < t < 5? 3) What is the acceleration at t = 3?

A particle moves on the x-axis with velocity at time t > 0 given as v(t) = -1 + e1 - t cm/min
4) Is the speed of the particle increasing at t = 3? Justify your answer. 5) What is the position of the particle at t = 5?

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