The position of a particle moving along an x axis is given...

The position of a particle moving along an $x$ axis is given by $x=12 t^{2}-2 t^{3}$, where $x$ is in meters and $t$ is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at $t=3.0 \mathrm{~s}$. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and $\underline{(g)}$ at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at $t=0$ )? (i) Determine the average velocity of the particle between $t=0$ and $t=\overline{3} \mathrm{~s}$

Key Concepts

Average Velocity

Average velocity is defined as the total change in position divided by the total time elapsed. It offers a macroscopic perspective on the particle's motion over a given time interval, smoothing out fluctuations in instantaneous velocity.

Acceleration Function and Its Computation

Acceleration is defined as the rate at which the velocity function changes with time, and it is found by taking the derivative of the velocity function. This gives insight into how the particle's motion is speeding up or slowing down.

Finding Extremum Points

In kinematics, finding extremum points involves determining where either the velocity or the position function attains its maximum or minimum values. This is typically done by setting the derivative(s) equal to zero and solving for the time instances, which indicate potential points of reversal or peak values in motion.

Position Function

This is the mathematical expression that represents the location of a particle as a function of time. It is fundamental in kinematics because it provides the starting point from which other aspects of motion—like velocity and acceleration—are derived.

Velocity Function and Its Computation

The velocity function is obtained by differentiating the position function with respect to time. It indicates the rate of change of position and provides information about both the speed and the direction of the particle's motion.

Transcript

00:01 In this problem of motion in a straight line, we have given that the position of a particle moving along x -axis is given by x is equals to 12 t squared minus 2 t cube where x is in meter and t is in seconds.

00:19 We have to determine the position, the velocity and the acceleration of the particle at t is equal to 3 seconds.

00:27 Now first we have to find the position at t is equal to 3 second.

00:33 Position that is x we have to plug in the value t is equal to 3 so x is equal to putting the value so this is 12 multiplied with 3 square minus 2 multiplied with 3 cube which is equals to 54 so we have 54 meters for part a now part b the velocity at 3 seconds so at first we have to find the velocity so when we derivative it, so by doing one time differentiation, this would become 24 and minus 8.

01:06 This is 2 multiplied with 3 which is 6 and t squared.

01:10 And now putting the value t is equal to 3.

01:12 So this is v3 is putting the value so this is 24 multiplied with 3 minus 6 multiplied with 3 square.

01:22 Now when we solve it this is equal to 18 meter per second.

01:27 Now part c we have to find the acceleration so that means we have to again differentiate vt so when we differentiate it so this is 24 minus 12 t now putting the value t is equal to 3 so a 3 is calculated as putting the value t is equal to 3 so this is 24 minus 12 multiplied with 3 so this is here now when we solve it this is equals to minus 12 meter per second square so this is the position at t is equal to 3 seconds and this is the velocity at t is equal to 3 second and this is the acceleration at t is equal to 3 second and now in part d.

02:08 Part d says what is the maximum positive coordinate reached by the particle? so the particle would be in motion when it has velocity that means this is velocity 2040 minus 60 square.

02:22 If this velocity is zero that means particle would be stopped.

02:25 So we have to put here 2040 minus 60 square.

02:29 Is square should be equal to 0 that means we have to find the value of t from here we calculate that t is equals to here this is 4 that means particle or we can say the position of moving particle would be varying at t is equal to 4 seconds now we have to find the position so putting so putting t is equal to 4 in the equation x is equal to 12 t square minus 2 t cube so here x is equal to 12 multiplied with 4 square minus 2 multiplied with 4 cube now when we solve it this is equal to 64 so that means this is the maximum coordinate reached by a particle that is 64 meter now part e at what time it is it so from here we have calculated at t is equal to 4 seconds so we have at t is equal to 4 second now part f part f says what is the maximum positive velocity that reached by the particle we have to find the maximum velocity reached by the particle so the velocity would be maximum when acceleration continues first it continues to accelerate then decelerate here we have distillation that means acceleration is minus so this term should be equals to 0 that means 24 minus 12 t should be equal to 0 from here we calculate at t is equal to 2 second at t is equal to 2 second the particle would have the maximum velocity so we have to plug in the value of velocity say velocity at t is equal to 2 second so velocity this is maximum putting the value so this is 24 multiplied with 2 which is 48 and this is 48 minus 6 multiplied with 2 square which is 24 so this is in meter per second so we have maximum velocity that is 24 meter per second...

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