Acceleration (a) If the position of a particle is given by...

Acceleration (a) If the position of a particle is given by $x=20 t-5 t^{3}$, where $x$ is in meters and $t$ is in seconds, when, is the particle's velocity zero? (b) When is its acceleration $a$ zero? (c) For what time range (positive or negative) is $a$ negative? (d) Positive? (e) Graph $x(t), v(t),$ and $a(t) .$

Acceleration
(a) If the position of a particle is given by $x=20 t-5 t^{3}$, where $x$ is in meters and $t$ is in seconds, when, is the particle's velocity zero? (b) When is its acceleration $a$ zero? (c) For what time range (positive or negative) is $a$ negative? (d) Positive? (e) Graph $x(t), v(t),$ and $a(t) .$

Key Concepts

Position Function

The position function defines the location of a particle along a line as a function of time. It is fundamental in kinematics because it gives the complete description of an object's location at any given moment, from which other motion attributes can be determined.

Velocity as the Derivative of Position

Velocity is obtained by differentiating the position function with respect to time. It measures how fast and in what direction the position is changing, serving as a bridge between position and acceleration.

Acceleration as the Derivative of Velocity

Acceleration is the derivative of the velocity function with respect to time, representing the rate at which velocity changes. It is a critical concept as it indicates how quickly an object's motion is speeding up or slowing down.

Finding Zeros of a Function

Finding the zeros of functions, such as those for velocity or acceleration, involves solving for the time values at which these derivatives equal zero. This concept is important for determining moments of rest or changes in motion dynamics.

Graphical Analysis of Motion

Graphing the position, velocity, and acceleration functions provides a visual depiction of motion. Such analysis helps in understanding the relationships between these kinematic quantities and how changes in one affect the others over time.

Transcript

00:01 Our exposition in terms of time would be equal to 20 t minus 5t cubed.

00:08 So essentially, our velocity in terms of time would be equal to 20 minus 15t squared.

00:15 And our acceleration in terms of time would be equal to negative 30t.

00:21 Now, for part a, we know that we can set the velocity equal to zero.

00:27 So that would be simply 20 minus 15 t squared equaling 0.

00:35 Essentially, t would be equal to only the positive value.

00:44 So t would be radical 20 over 15.

00:48 And this is equaling 1 .2 seconds.

00:52 And we can say for part b sets the acceleration equal to 0, this would be negative 30 t and so at t equals zero seconds is when the acceleration would equal 0 meters per second squared for part e rather for my apologies for part c we can say that it is clear that a of t equaling negative 30 t this is going to be negative for all t being greater than zero and then for part d we know that of course a of t equaling negative 30 t this would be positive for all t being less than zero so we can say for part e if we want to draw the rather we can just sketch the position versus time this would be our x position in meters and it would if again we have to say that this is going to be approximately 1 .2 seconds.

02:28 So the sketch would essentially be something like this.

02:37 Maybe this would be a little bit smoother.

02:42 And again, right here, it's 1 .2 seconds because that's when we know that it's going to cross the origin.

02:48 And then we can say that, oh, actually, my apologies, this would actually prove to be, this would actually prove to be where the x crosses the origin.

03:05 So we'd actually have to solve this equation...

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