The position of a particle moving along the x axis depends...

The position of a particle moving along the $x$ axis depends on the time according to the equation $x=c t^{2}-b t^{3}$, where $x$ is in meters and $t$ in seconds. What are the units of (a) constant $c$ and (b) constant $b$ ? Let their numerical values be $3.0$ and $2.0$, respectively. (c) At what time does the particle reach its maximum positive $x$ position? From $t=0.0 \mathrm{~s}$ to $t=4.0 \mathrm{~s},(\mathrm{~d})$ what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) $1.0 \mathrm{~s}$, (g) $2.0 \mathrm{~s}$, (h) $3.0 \mathrm{~s}$, and (i) $4.0 \mathrm{~s}$. Find its acceleration at times (j) $1.0 \mathrm{~s}$, (k) $2.0 \mathrm{~s},(1) 3.0 \mathrm{~s}$, and $(\mathrm{m}) 4.0 \mathrm{~s}$.

The position of a particle moving along the $x$ axis depends on the time according to the equation $x=c t^{2}-b t^{3}$, where $x$ is in meters and $t$ in seconds. What are the units of (a) constant $c$ and (b) constant $b$ ? Let their numerical values be $3.0$ and $2.0$, respectively. (c) At what time does the particle reach its maximum positive $x$ position? From $t=0.0 \mathrm{~s}$ to $t=4.0 \mathrm{~s},(\mathrm{~d})$ what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) $1.0 \mathrm{~s}$,
(g) $2.0 \mathrm{~s}$, (h) $3.0 \mathrm{~s}$, and
(i) $4.0 \mathrm{~s}$. Find its acceleration at times
(j) $1.0 \mathrm{~s}$,
(k) $2.0 \mathrm{~s},(1) 3.0 \mathrm{~s}$, and $(\mathrm{m}) 4.0 \mathrm{~s}$.

Key Concepts

Kinematic Equations and Constant Determination

Kinematic equations describe the motion of objects in terms of displacement, velocity, and acceleration with respect to time. When these equations include constants, it is imperative to determine the units and roles of these constants through methods such as dimensional analysis, ensuring that the equations accurately reflect physical reality.

Displacement versus Distance

Displacement is a vector quantity that represents the net change in position of an object, while distance is a scalar quantity that represents the total length of the path traveled. In kinematic problems, it is important to distinguish between these two concepts when analyzing motion over a given time interval.

Extremum Conditions in Motion

Determining maximum or minimum positions in motion involves finding the extremum points of the position function. This is typically done by setting the first derivative (velocity) to zero and solving for time. Once these points are identified, further analysis, such as the second derivative test, can confirm whether they are maxima or minima.

Differentiation in Kinematics

Differentiation is a mathematical tool used in kinematics to relate position, velocity, and acceleration. By taking the first derivative of a position function with respect to time, one obtains the velocity function, and a second derivative yields the acceleration function. This process is essential in analyzing how an object’s motion changes over time.

Dimensional Analysis

This concept involves ensuring that every term in a physical equation has consistent units. When given an equation with constants and variables, dimensional analysis is used to determine the necessary units of the constants so that the equation remains valid in terms of physical dimensions, such as length and time.

Transcript

0:00 Okay.

00:02 So now the position of a particle is x equals ct squared minus bt cubed.

00:20 What are the units of c and b? and b well the units of c because it needs to be meters in the end but then you're multiplying by second squared that needs to be meters per second squared b needs to be meters per second cubed because we're multiplying by t to the third power okay we're going to let a equal three and b equal i mean c d uh get my b and c mixed up? oh, no, no, i didn't.

01:05 Okay, c is three and b is two.

01:10 So now i'm going to erase this.

01:14 C is three and b is two.

01:19 Okay.

01:24 At what time does it reach its maximum position? well, if i calculate the velocity using a derivative, that would be 6t minus 6t squared.

01:42 And his max position is going to be when the velocity equals zero.

01:48 So that's going to be 0 equals 6t minus 6t squared.

01:53 So dividing by the 6 cancels out.

01:56 So 0 is going to equal t times 1 minus t.

02:06 So then i get t equals 0 and t equals 1.

02:12 One of these is a max and the other one is a min.

02:16 So i'm going to go to the calculator and i'm going to type in 3t squared.

02:30 Well, i'm going to put in t equals zero first.

02:33 Then i'm going to type in 3t squared minus 2t cubed.

02:42 At zero is zero.

02:45 At 1, it's 1.

02:50 So the max is at t equals 1.

02:55 The min is at t equals 0.

02:57 So c, from t, so that's 1, from t equals 0 to t equals 4, what distance does the particle move? okay, so d, i just need to take x of 4 minus x of 0.

03:23 Well, x of 0 is 0.

03:24 So i really just need to get x of 4 negative 80...

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