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

The position of a particle moving along the $x$ axis is given in centimeters by $x=9.75+1.50 x^{3},$ where $t$ is in seconds. Calculate (a) the average velocity during the time interval $t=2.00 \mathrm{~s}$ to $t=3.00 \mathrm{~s}$ (b) the instantaneous velocity at $t=2.00 \mathrm{~s} ;$ (c) the instantaneous velocity at $t=3.00 \mathrm{~s} ;$ (d) the instantancous velocity at $t=2.50 \mathrm{~s} ;$ and (e) the instantaneous velocity when the particle is midway between its positions at $\ell=2.00 \mathrm{~s}$ and $t=3.00 \mathrm{~s}$. ( $($ ) Graph $x$ versus $t$ and indicate your answers graphically.

The position of a particle moving along the $x$ axis is given in centimeters by $x=9.75+1.50 x^{3},$ where $t$ is in seconds. Calculate (a) the average velocity during the time interval $t=2.00 \mathrm{~s}$ to $t=3.00 \mathrm{~s}$
(b) the instantaneous velocity at $t=2.00 \mathrm{~s} ;$ (c) the instantaneous velocity at $t=3.00 \mathrm{~s} ;$ (d) the instantancous velocity at $t=2.50 \mathrm{~s} ;$ and
(e) the instantaneous velocity when the particle is midway between its positions at $\ell=2.00 \mathrm{~s}$ and $t=3.00 \mathrm{~s}$. ( $($ ) Graph $x$ versus $t$ and indicate your answers graphically.

Key Concepts

Average Velocity

Average velocity is defined as the change in position divided by the time interval over which that change occurs. It provides a measure of the overall rate of displacement during a finite interval and is calculated using the formula ?x/?t.

Instantaneous Velocity

Instantaneous velocity describes the rate of change of position at a specific moment in time. It is found by taking the derivative of the position function with respect to time, which gives the slope of the tangent to the position-time graph at a particular point.

Derivative and Differentiation

Differentiation is a fundamental concept in calculus that quantifies how a function changes as its input changes. In kinematics, taking the derivative of a position function with respect to time yields the velocity function, providing a precise measure of instantaneous velocity.

Position-Time Graph

A position-time graph visually represents the movement of an object with time on the horizontal axis and position on the vertical axis. The slope of the graph at any point indicates the instantaneous velocity, while the average slope over an interval indicates the average velocity between those times.

Transcript

00:01 In this problem of motion along a straight line we have given that the position of a particle moving along x -axis is given in centimeters.

00:10 So this is x is in centimeters and this is the equation that is 9 .75 plus 1 .50 t cube.

00:24 Where t is in seconds.

00:26 First we have to calculate the average velocity during the time interval t is equal to 2 second to 3 seconds.

00:34 T is equal to 2 .00 seconds to t is equal to 3 .00 seconds so for this first we have to find the position at t is equal to 2 seconds so we see this is the position at t is equal to 2 second that is x2 now putting t is equal to 2 so this is 9 .75 plus 1 .50 multiplied with 2 q which is 8 now when we solve it this is equal to 21 .75 this is centimeter now position at t is equal to 3 second that is x3 putting t is equals to 3 so this is 1 .5 this is 1 .50 multiplied with 2 this is here we have to place 3 cube so 27 so multiplied with 27 x3 is equal to 9 .75 plus 1 .50 multiplied 25 is equals to 50 .25 cm.

01:42 Now we have to find the average velocity, so average velocity v average is equal to change in displacement say x3 minus x2, say x3 minus x2 divided with the time taken say 3 minus 2 second that is equals to 1 second.

02:01 So when we put the value 50 .25 minus 21 .75 so this is equal to 28.

02:09 5 cm per second.

02:12 So for answer we have to write here 28 .5 centimeter per second.

02:19 This is the answer for part a.

02:21 Now part b says the instantaneous velocity at t is equal to 2 second.

02:30 We have to find the instantaneous velocity as 2 .00 seconds.

02:34 So for this instantaneous velocity say v instantaneous is given by dx divided with dt.

02:41 That means we have to differentiate this term so when we differentiate it so this value would be 9 .75 differentiation of constant is equal to 0 so this is 0 plus 1 .50 multiplied with 3 so this is 4 .50 multiplied with t squared now we want the instantaneous velocity at t is equal to 2 seconds so we have to put t is equal to 2 so v instantaneous at 2 second is equal to 2 is equal to 2 is equal to 4 .5 5 .50 multiplied with 2 square which is equal to 4 .50 multiplied with 4 is equal to 18 cm per second...

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