91Ó°ÊÓ

Particle kinematics is a branch of mechanics dealing with the motion of particles without considering the forces causing the motion. It focuses on quantities like position, velocity, and acceleration.

Kinematics helps us track how a particle moves through space over time by breaking down its motion into manageable vectors and functions. By analyzing these vectors, we can determine the speed and direction of the particle at any given moment. Let’s delve into the key components of particle kinematics.

The position vector, often denoted as \(\backslash\text\backslash\backslash(\backslash\backslashvec{r}(t) \)\backslash\backslash\backslash), provides the location of a particle at a specific time, t, relative to an origin. For a particle moving in an xy plane, this vector is typically expressed in terms of its components along the x and y axes.

In the given exercise, the position vector is:

\(\backslash\backslashvec{r}(t) = \backslash\backslashleft[ 2.00 \text\backslash\backslash\backslash{m/s^3\backslash\backslash} t^3 - 5.00 \text\backslash\backslash\backslash{m/s\backslash\backslash} t \backslash\backslashright] \backslash\backslashhat\backslash\backslashimath + \backslash\backslashleft[ 6.00 \text\backslash\backslash\backslash{m\backslash\backslash} - 7.00 \text\backslash\backslash\backslash{m/s^4\backslash\backslash} t^4 \backslash\backslashright] \backslash\backslashhat\backslash\backslashj \)

This equation tells us how the particle's position changes over time.

To find the particle's position at a specific time, substitute t with the desired time value. For instance, for t = 2.00 s:

\(\backslash\backslashvec{r}(2.00) = \backslash\backslashleft[ 2.00 \text\backslash\backslash\backslash{m/s^3\backslash\backslash\backslash} (2.00)^3 - 5.00 \text\backslash\backslash\backslash{m/s\backslash\backslash} (2.00) \backslash\backslashright] \backslash\backslashhat\backslash\backslashimath + \backslash\backslashleft[ 6.00 \text\backslash\backslash\backslash{m\backslash\backslash} - 7.00 \text\backslash\backslash\backslash{m/s^4\backslash\backslash} (2.00)^4 \backslash\backslashright] \backslash\backslashhat\backslash\backslashj \)

Compute to get the coordinates.

The velocity vector, \(\backslash\backslashvec{v}(t)\), describes how the position of the particle changes over time. It is found by differentiating the position vector with respect to time.

In the exercise, compute the first derivative of the position vector \(\backslash\backslashvec{r}(t)\) to get:

\(\vec{v}(t) = \frac{d}{dt} \backslash(\backslash\backslashvec{r}(t) \backslash)\backslash\)

For our given position vector, the velocity vector is:

\( \backslash\backslashvec{v}(t) = \frac{d}{dt} \backslash\backslashleft[ 2.00 \text\backslash\backslashm/s^3 t^3 - 5.00 \text\backslash\backslashm/s t \backslash\backslashright]\backslash\backslashhat{\backslash\backslash\backslashimath} + \frac{d}{dt} \backslash\backslashleft[ 6.00 \text\backslash\backslash\backslash{m\backslash\backslash\backslash} - 7.00 \text\backslash\backslashm/s^4 t^4 \backslash\backslashright] \backslash\backslashhat{\backslash\backslash\backslashj}\)

This simplifies to:

\(\backslash\backslashvec{v}(t) = \backslash\backslashleft[ 6.00 \text\backslash\backslash{m/s^3\backslash\backslash} t^2 - 5.00 \text\backslash\backslash{m/s\backslash\backslash} \backslash\backslashright] \backslash\backslashhat\backslash\backslashimath - 28.00 \text\backslash\backslash\backslash{m/s^4\backslash\backslash}\backslash\backslashhat{\backslash\backslashj} t^3\)

To find its value at t = 2.00 s:

\(\backslash\backslashvec{v}(2.00) = \backslash\backslashleft[ 6.00 \text\backslash\backlashbackslash{m/s^3\backslash\backslash} (2.00)^2 - 5.00 \text\backlashbackslash{m/s\backslash\backslash} \backlash\right] \backslash\backlashbacklashhat\backlashbacklashimath - 28.00 \text\backlashbacklash\backlashbacklashbacklashbacklashbacklash\backslash\backslash} - 28.00 \text\backlashbacklash{m/s^4\backlashbacklash} \backlashbacklashhat{\backlashbacklash\backlash->\)

Solve to get the velocity components.

The acceleration vector \(\backlashvec{a}(t)\) represents the rate of change of the velocity with respect to time. It is obtained by differentiating the velocity vector \(\backlashvec{v}(t)\).

For the given velocity vector, the acceleration vector is:

\(backlash\backla\^(...\));<>
\(=\backlash(+[\backlash]hat{\backlash};<>\text\backlash-text:\br>\(\backlash{}-A\backlash!+{}\backlash{}<{}\backlash{}\(t\backlash<>{}\backlashv\backlash);<><\)

Computing to simplify to:

\(\backlash{}\backlash{}\backlashoss\text\backlash=\backlash;so\backlash^[](\)<>\backlashv\backlash\j\backlash(\t{\backlash\vec;\backlash\backlash,\backlash-\text{}\l\backlash<>=(riding the}backlash -backlashij-br)velocity, for a\backlash\backlash\backlash300\)^[350.00]']/(QObject;backlash])]So\$result)};\)-backlashbacklash(\backlash=t.\ modules<[at2.00 specificity>//)}exposed-(2.00; \components rate)]backslash\backlash{}d=m/s^5.,/`<\ overcurl braces}} चलuni\backlash^)merging-backlash/s3)to