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. (a) What units must $c$ and $b$ have? Let their numerical values be $3.0$ and $2.0 .$ respectively. (b) At what time does the particle reach its maximum positive $x$ position? From $t_{0}=0.0 \mathrm{~s}$ to $t_{4}=4.0 \mathrm{~s},(\mathrm{c})$ what distance does the particle move and (d) what is its displacement? At $t=1.0$, $2.0,3.0$, and $4.0 \mathrm{~s}$, what are (e) its velocities and (f) its accelerations?

Key Concepts

Distance Versus Displacement

Distance and displacement are two distinct concepts in kinematics. Displacement refers to the change in position from the starting point to the end point and is a vector quantity that includes direction. Distance, on the other hand, is a scalar quantity representing the total length of the path traveled, independent of direction.

Finding Extremum Points in Motion

In problems involving motion, determining the instant at which a particle reaches a maximum or minimum position involves finding the extremum points of the position function. This is typically achieved by setting the first derivative (velocity) equal to zero and analyzing the sign changes or using the second derivative test to confirm whether the point is a maximum or minimum.

Differentiation in Kinematics

Differentiation is a fundamental mathematical tool used in kinematics to extract velocity and acceleration from the position-time relationship. The first derivative of position with respect to time gives instantaneous velocity, while the second derivative provides instantaneous acceleration, allowing one to analyze the dynamic behavior of moving particles.

One?Dimensional Kinematics

One-dimensional kinematics deals with the motion of particles along a straight line. It involves understanding how position, velocity, and acceleration are interrelated over time and solving problems that require analyzing the particle's displacement, distance traveled, and behavior at special points like maxima or minima of its trajectory.

Dimensional Analysis

Dimensional analysis is used to determine the appropriate physical units for constants in equations. This process involves matching the dimensions on both sides of an equation to ensure consistency, a crucial step in verifying equations in physics, especially in kinematics where position, time, and acceleration are related.

Transcript

00:01 The motion of the position of the particle x equals to given for me for us that x equals to ct square minus b tq okay so c t square minus b t qu this is the motion so first of all you need to find the unit so basically x is in the distance that means meter so the unit of a will be it should be unit of the meter so that means the meter obviously length that means the meter so it should be unit of the c that should be nothing but meter per second square because second square and the second square cancel out and this meter will be remaining out and b unit will be a thing of meter per second q okay in the next scenario we have given that b c that c and b having value 3 and 2 right so c equals to 3 and b equals to 2 so now we need to find that we need to find what it was time does the particle reaches its maximum position x 0 to 4 second okay so basically now x becomes nothing but the c t squared so 3 t squared minus 2 t q okay so you need to find at 0 to 4 second at what point this becomes the maximum so you can say when we put 0 this will be 0 when we put 1 this will be 2 3 when you put 2 this will be 12 minus of uh 12 minus of so we can just calculate the maximum.

01:40 Where x is maximum, the xx by d equals to 0, right? so it means 6 t minus 3 t minus of 6 t squared equals to 0.

01:54 So t equals to 1.

01:56 So obviously at t equal to 1, the particle will be at maximum position of the x -axis.

02:02 In the next part, i want to find that, but we need to find that what distance does the particle move and so 0 to 4 second we know that for 0 to 4 second what distance but does the particle move okay so how can we find that so we know that first of all we need to find the acceleration and their velocities okay or d x upon at equal to 0 obviously x equals 2 is at 0 and first of all we will just we will just find the positions and after that you can also find so at t equal to 0 it is at x equal to 0 okay and at t equal to 1 it is at x equal to 1 okay so that means it is moving from a 0 to 1 right and then at x equal to 4 it is 16 and 48 16 and 3 so 48 it minus of 16 1128 so it will be a t minus of 80 okay so now finally initially it is at the position of 0 and then it comes to x equal to 1 and this is finally it is going to minus 80 so how are the distance this is traveling 1 and plus 1 2 and plus 80 central distance d equal to 82 by the formula of the graph okay in the next part what we need to find is that what is the displacement at x t equal to 1 so obviously at t equal to 1 x equal to 1 and the initial and function is the 1 so a displacement s 1 is equal to 1 obviously and at equal to 0 obviously at t equal to 2 what is their position so this will be 2 into 2 4 12 and minus 2 into 2 2 2 2 2 2 4 2 or 21 16 so it will be of 4 right so what will be the displacement so initially going from over to over and at the minus 4 so now the displeasement s 2 will be minus of 4 and s equal to 3 so at s equal to 3 it will be nothing but 9 into 3 so that will be 27 minus of 4 so this will be minus of 27 s 3 equal to minus of 27 and obviously s 4 is t equal to 4 that means minus of 40 which we have already calculated now the next part we have to calculate that its velocity is and its acceleration so velocity v equals to nothing but for it velocity will differentiate so it will be 60 minus of 60 square v equals to 60 minus of 60 square so as we at one second v equal to 0 at 2 second it will be 12 minus of for 24 so minus of 12 meter per second and v3 at 3 equal to 3 that means 18 minus of 9 into 6 so 36 so it will be 18 and v 4 this will 24 that will be 40 meter per second right now in the next part we need to find the acceleration so acceleration a equals to dv by d t so 60 means 6 minus of 6 t squared so 12 t okay so obviously acceleration at 1 second will be nothing but minus of 6 meter per second square...

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