The position of a particle between t=0 and t=2.00 s is given by x(t)=(3.00 m / s^3) t^3-(10.0 m

The position of a particle between t=0 and t=2.00 s is given...

The position of a particle between $t=0$ and $t=2.00 \mathrm{s}$ is given by $x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t$ (a) Draw the $x-t, v_{x}-t,$ and $a_{x}-t$ graphs of this particle. (b) At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle instantaneously at rest? Does your numerical result agree with the $v_{x}-t$ graph in part $(a) ?$ At each time calculated in part $(b)$ is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from $a_{x}(t)$ and from the $v_{x}-t$ graph. (d) At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the velocity of the particle instantancously not changing? Locate this point on the $v_{x} t$ and $a_{x}-t$ graphs of part (a). (e) What is the particle's greatest distance from the origin $(x=0)$ between $t=0$ and $t=2.00 \mathrm{s} ?(\mathrm{f})$ At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle speeding $u p$ at the greatest rate? At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle slowing down at the greatest rate? Locate these points on the $v_{x}-t$ and $a_{x}-t$ graphs of part (a).

The position of a particle between $t=0$ and $t=2.00 \mathrm{s}$ is given by $x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t$ (a) Draw the $x-t, v_{x}-t,$ and $a_{x}-t$ graphs of this particle. (b) At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle instantaneously at rest? Does your numerical result agree with the $v_{x}-t$ graph in part $(a) ?$ At each time calculated in part $(b)$ is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from $a_{x}(t)$ and from the $v_{x}-t$ graph. (d) At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the velocity of the particle instantancously not changing? Locate this point on the $v_{x} t$ and $a_{x}-t$ graphs of part (a). (e) What is the particle's greatest distance from the origin $(x=0)$ between $t=0$ and $t=2.00 \mathrm{s} ?(\mathrm{f})$ At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle speeding $u p$ at the greatest rate? At what time(s) between $t=0$ and $t=2.00 \mathrm{s}$ is the particle slowing down at the greatest
rate? Locate these points on the $v_{x}-t$ and $a_{x}-t$ graphs of part (a).

Key Concepts

Derivatives in Kinematics

This concept involves using calculus to relate position, velocity, and acceleration. The velocity of a particle is obtained as the first derivative of the position function with respect to time, and the acceleration is the second derivative (or the derivative of the velocity function). This process is essential for analyzing the particle’s motion, as it converts a position-time function into meaningful insights about speed and changes in speed.

Graphical Analysis of Motion

Understanding and interpreting the x-t, v-t, and a-t graphs is a fundamental skill in kinematics. These graphs provide information about the particle's displacement, instantaneous velocity, and acceleration respectively. Analyzing these graphs helps in visualizing the behavior of the particle over time, identifying key events like changes in direction, points of rest, and intervals of speeding up or slowing down.

Instantaneous Rest

This key concept refers to moments when the velocity of the particle is exactly zero. Identifying the time(s) at which velocity becomes zero is critical for determining when a particle is momentarily at rest, even if other aspects of its motion (like acceleration) are nonzero. This condition is found by solving for the roots of the velocity function.

Critical Points and Extremum Analysis

Critical points in the motion of a particle, such as maximum displacement or zero acceleration points, are determined by analyzing where the derivative (velocity or acceleration) equals zero or changes sign. This method is used to locate the time(s) at which a particle reaches its maximum or minimum position relative to the origin, and to determine when the rate of change of velocity is nil.

Speeding Up and Slowing Down

This concept involves understanding the relationship between the signs of velocity and acceleration. A particle speeds up when velocity and acceleration have the same sign (both positive or both negative) and slows down when they have opposite signs. Correct interpretation of these signs is crucial for analyzing the dynamics of the particle’s motion, such as identifying when the particle is rapidly increasing or decreasing its speed.

Transcript

00:01 In this question we are given this particular equation that describes the position of a particle.

00:07 And we want to draw the graph for the position time as well as the velocity time and acceleration time graph.

00:16 So first of all what you need to do is to get the velocity equation.

00:21 We can do that by just differentiating the position equation with respect to time.

00:28 Right and we should get 9 t square minus 20 t plus 9 so this very simple differentiation just differentiating each term and then of course to get the acceleration term we just differentiate our velocity equation right so now we have these three equations we can start to draw our graphs or to sketch the graphs.

01:17 So over here, the first one, the position one, we know that, sorry, i'm going to start up with the easiest one, which is the acceleration curve.

01:27 For the acceleration curve, we see that it's just a linear straight line.

01:33 We have starting point, right, when t is equal to zero, starting point is a negative value, but it has a positive gradient.

01:43 So a positive gradient starting from negative 20.

01:49 So expect something like this, cutting through the t -axis.

02:02 On the other hand for the velocity, we see that it starts off at positive value, 9.

02:11 And because it is a quadratic equation, a positive one, the term t -square has a positive factor coefficient.

02:23 It means that it is a u -shaped quadratic curve.

02:33 So starting from 9, we expect it to be something like this.

02:45 For the position, the position curve it will be a cubic equation, it's a cubic equation so it will have a minimum and a maximum point.

03:03 And the starting position when t is equals to 0 is 0.

03:10 Start off at 0.

03:13 And because the initial velocity is positive, we expect it to have a positive gradient and it decreases to negative.

03:23 So we expect the gradient to go negative.

03:32 So gradient decreases to 0 and negative.

03:37 Then it goes back up again to positive.

03:43 So something like that.

03:45 This is just a rough sketch.

03:48 Right so now we want to find some numerical answers such as the time between zero and two seconds.

03:58 You want to find when the particle is instantaneously at rest.

04:03 So being instantaneously at rest means that our velocity is zero.

04:08 So we can solve for this quadratic equation by grading this to 0 using the quadratic formula negative b plus minus b square where a is 9 b is negative 20 and c is 9 so we should get t to b equals to 0 .63 and 1 .59 seconds so these two results.

05:05 And from the curve, if you were to draw an exact curve, this should be the two points, which cut the t -axis at 0 .63 and also 1 .59.

05:28 Now for the third part of the question, i'm going to find when the if the acceleration at these two different points, points is positive or negative.

05:47 So depending on the tangent line, the gradient line at these two points actually determine the acceleration, but it's positive or negative...

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