5. An object moving in the x-y plane has a position vector given by: 7 = (2t³ - 5t)i + (6 - t¹)ĵ (in meters). nd the particle's a. position vector at t = Os b. position vector at t = 2s 4 c. displacement vector for this change in position d. average velocity during this motion
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
Only d
![### Vector Motion in the x-y Plane
Consider an object moving in the x-y plane with a position vector represented by:
\[ \vec{r} = (2t^3 - 5t) \hat{i} + (6 - t^4) \hat{j} \]
where \( \vec{r} \) is in meters, and \( t \) is in seconds.
To analyze the motion of the particle, solve the following:
**a. Position vector at \( t = 0 \)s**
\[ \vec{r}(t=0) = (2(0)^3 - 5(0)) \hat{i} + (6 - (0)^4) \hat{j} \]
\[ \vec{r}(0) = 0 \hat{i} + 6 \hat{j} \]
\[ \vec{r}(0) = 6 \hat{j} \text{ meters} \]
**b. Position vector at \( t = 2 \)s**
\[ \vec{r}(t=2) = (2(2)^3 - 5(2)) \hat{i} + (6 - (2)^4) \hat{j} \]
\[ \vec{r}(2) = (2(8) - 10) \hat{i} + (6 - 16) \hat{j} \]
\[ \vec{r}(2) = (16 - 10) \hat{i} + (-10) \hat{j} \]
\[ \vec{r}(2) = 6 \hat{i} - 10 \hat{j} \text{ meters} \]
**c. Displacement vector for this change in position**
The displacement vector is the difference in the position vectors at \( t=2 \)s and \( t=0 \)s:
\[ \Delta \vec{r} = \vec{r}(2) - \vec{r}(0) \]
\[ \Delta \vec{r} = (6 \hat{i} - 10 \hat{j}) - (0 \hat{i} + 6 \hat{j}) \]
\[ \Delta \vec{r} = 6 \hat{i} - 16 \hat{j} \text{ meters} \]
**d. Average velocity during this motion**
The](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbad9eda7-57c9-475e-b126-0f24ea0e5433%2F764fc4aa-6f0e-4154-92eb-4eeac2e27ea8%2Fy8st9bu_processed.jpeg&w=3840&q=75)

Step by step
Solved in 2 steps









