The position function x(t) of a particle moving along an x...

The position function $x(t)$ of a particle moving along an $x$ axis is $x=4.0-6.0 u^{2},$ with $x$ in meters and $t$ in seconds. (a) $\Lambda t$ what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph $x$ versus $t$ for the range $-5 \mathrm{~s}$ to $+5 \mathrm{~s}$ (f) To shift the curve rightward on the graph, should we include the term $+20 t$ or the term $-20 t$ in $x(t) ?($ g) Docs that inclusion increase or decrease the value of $x$ at which the particle momentarily stops?

Key Concepts

Function Translation and Shifting

Translating or shifting a function involves altering its input or output to change the position of its graph. Understanding how adding or subtracting a term, such as a linear term in time, affects the curve is crucial for modifying the function to achieve desired shifts in its graphical representation.

Graphical Analysis and Interpretation

Graphing the position function versus time provides a visual representation of the particle’s motion. This analytical tool helps in understanding the behavior of the function over a range, identifying key features such as stops, crossings, and trends in the motion.

Velocity as the Derivative of Position

Velocity is defined as the time derivative of the position function. Finding when the velocity is zero is essential for identifying moments when the particle stops momentarily, which helps in analyzing the motion’s turning points and overall dynamics.

Position Function

The position function describes the spatial location of a particle as a function of time. It encapsulates all the information required to understand the particle's motion along a one?dimensional path, allowing us to determine where the particle is at any given moment.

Root Finding in Polynomial Equations

Solving for the roots of an equation, such as determining the times when the position function equals zero, is a vital skill. This involves manipulating polynomial equations to find the specific instances in time at which a particle reaches a particular position.

Transcript

00:01 In this problem of motion along a straight line we have given that the position function xt, this is position function xt of a particle moving along x -axis is, say x is equals to 4 .0 minus 6 .0 t square where x is measured in meters and t is measured in seconds.

00:25 Now it is asked that at what time and where does the particle momentarily? it stops.

00:34 So this particle would stop when the velocity is equal to 0.

00:37 So dv, that is here dx divided with d t.

00:41 So here v is equal to dx divide with t would be equal to zero.

00:46 So for this we have to differentiate it.

00:48 Now when we differentiate it, so this value would be minus 12.

00:52 So this is minus 12 multiplied with t.

00:59 That means particle momentarily stops when minus 12 multiplied with t is equal to 0 that means at t is equal to 0 this particle momentarily it stops so at say time t is equal to 0 second it is stops and it is at the origin now this would be not at origin because when we put t is equal to 0 the position would be 4 .0 so we can say at 4 .0 meters and now it is asked that at what positive and negative time the particle pass through origin so first we have to tell the negative time and then in part d we have to tell the positive so for this we have to put x is equals to 0 by putting x is equal to 0 so this value would be 4 .0 minus 6 .0 t 2 .0 2 .2 so 2 is so t is square is equal to 4 divided with 6 that is 2 divided with 3 now t is equal to under root 2 divided with 3 3.

02:13 This is equals to here t is equals to plus and minus 0 .816 second...

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