The position of an object moving along an x axis is given by...

The position of an object moving along an $x$ axis is given by $x=3 t-4 t^{2}+t^{3},$ where $x$ is in meters and $t$ in seconds. Find the position of the object at the following values of $t:$ (a) $1 \mathrm{~s}$. (b) $2 \mathrm{~s}$. (c) $3 \mathrm{~s}$, and (d) $4 \mathrm{~s}$. (c) What is the object's dicplacement hetween $t=0$ and $t=4 \mathrm{~s} ?$ (f) What is its average velocity for the time interval from $t=2 \mathrm{~s}$ to $t=4 \mathrm{~s} ?$ (g) Graph $x$ versus $t$ for $0 \leq t \leq 4 \mathrm{~s}$ and indicate how the answer for (f) can be found on the graph.

The position of an object moving along an $x$ axis is given by $x=3 t-4 t^{2}+t^{3},$ where $x$ is in meters and $t$ in seconds. Find the position of the object at the following values of $t:$ (a) $1 \mathrm{~s}$. (b) $2 \mathrm{~s}$.
(c) $3 \mathrm{~s}$, and (d) $4 \mathrm{~s}$. (c) What is the object's dicplacement hetween $t=0$ and $t=4 \mathrm{~s} ?$ (f) What is its average velocity for the time interval from $t=2 \mathrm{~s}$ to $t=4 \mathrm{~s} ?$ (g) Graph $x$ versus $t$ for $0 \leq t \leq 4 \mathrm{~s}$ and indicate how the answer for (f) can be found on the graph.

Key Concepts

Graphical Analysis of Motion

Graphical analysis in kinematics involves plotting the position function versus time to visualize the motion of an object. Features such as the slope of the position-time graph can be used to interpret the instantaneous velocity, while the slope of a secant line between two points on the graph yields the average velocity over that time interval.

Average Velocity

Average velocity is defined as the total displacement divided by the time interval during which the displacement occurred. It provides a measure of how fast an object is moving on average over that interval and is a vector quantity, meaning it also includes direction.

Displacement

Displacement is defined as the net change in position of an object over a given time interval. It is obtained by subtracting the initial position from the final position. This quantity captures the overall effect of the movement in a specific direction, regardless of the actual path taken.

Position Function

A position function represents how the spatial location of an object changes with time. In kinematics, it is typically expressed as a function x(t), which describes the object's motion along a specified axis. This function allows one to determine where the object is located at any given instant by substituting different time values.

Evaluation of the Position Function

Evaluating the position function at specific times involves substituting particular time values into the function to find the corresponding positional coordinates. This process is essential for understanding the object’s state of motion at different moments and lays the foundation for further analyses like determining displacement or velocity.

Transcript

00:01 In this problem of motion along a straight line we have given that the position of an object moving along x -axis is given why x is equals to 3 t minus 4 t square plus t cube and now where x is measured in meters so x we have measured in meters and t is measured in seconds we have to find the position of the object at t is equal to first one second, then t is equals to two second, then t is equal to three second and then t is equal to four second.

00:45 So first we have to solve a, b c and d parts of for this.

00:50 For solving the position of the object at t is equal to one second we have to put t is equal to one in this equation say this is equation number 1 now putting t is equal to 1 so x is equal to 3 multiplied with 1 which is 3 minus 4 multiplied with 1 square which is 4 plus 1 cube is again 1 so this is equal to 0 meters at t is equal to 2 second we have to put t is equal to 2 in this equation so 3 multiplied with 2 is equal to 6 minus 4 multiplied with 2 square which is 4 so 4 multiplied with 4 is equals to 16 and 2 cube so 2 cube is equal to 8 now when we solve it this is equals to 8 plus 6 14 minus 16 is equal to minus 2 meters now position of the object at t is equal to 3 second so for this we have to put t is equal to 3 in this equation 3 multiplied with 3 which is 9 minus 4 multiplied with 3 square which is 9 that is 36 plus 3 cube is equal to 27 now when we solve it this is equal to 27 plus 9 that is 36 minus 36 is equal to 0 meters.

02:01 Now we have to find the position of the object at t is equal to 4 seconds.

02:04 So for this we have to put t is equal to 4 in this equation.

02:07 Now putting t is equal to 4.

02:08 So 3 multiplied with 4 is 12 minus 4 multiplied with 4 square which is 4 square is 16.

02:16 So this is 16 plus 4 cube is 64.

02:21 Now when we solve it this is equal to 2.

02:25 Now we have to find the part e.

02:30 In part e we have asked that what is the object's displacement between t is equal to 0 and t is equal to 4 seconds...