Question

The position of an object moving along an x axis is given by,$x=3t-4t^2+t^3$ where x is in metres and tin seconds. Find the position of the object at following values of t:a) 1sec,(b)2sec, (c) 3sec, and (d) 4sec,(e) What is the object’s displacement between t=0and t=4sec? (f)What is its average velocity for the time interval from t=2sto t=4s? (g)Graph x vs t for$0\le t\le 4sec$and indicate how the answer for f can be found on the graph.

Short answer

1. The position of the object at t=1s is X=0m.
2. The position of the object at t=2s is X=-2.
3. The position of the object at t=3s is X=0.
4. The position of the object at t=4s is X=12.
5. The displacement of the object is +12m.
6. Average velocity of the object is +7m/s.

Step-by-step solution

Given data

The position of the object,$x\left(t\right)=3t-4t^2+t^3$

Understanding the displacement and average velocity

Displacement of a particle gives the change in position of the particle. Average velocity is the ratio between the total displacement to the total time interval.

The expression for the displacement is as follows:

$\Delta x=x_2-x_1$ (i)

Here, $X_1$is the initial position, and is $X_2$the final position.

The expression for the average velocity is as follows:

$v_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}$ (ii)

Here, $\Delta X$is the displacement, and $\Delta t$is the time interval.

(a) Determination of the position at t = 1 s

The position of the object is given as follows:

$x\left(t\right)=3t-4t^2+t^3$

The position of the object at t=1 s is calculated as follows:

$\begin{array}{c}x\left(1\right)=3\left(1\right)-4\left(1^2\right)+\left(1^3\right)\\ =4-4\\ =0\mathrm{m}\end{array}$

Thus, the position of the object at t=1 s is 0 m.

(b) Determination of the position at t = 2 s

The position of the object at t=2 s is calculated as follows:

$\begin{array}{c}x\left(2\right)=3\left(2\right)-4\left(2^2\right)+\left(2^3\right)\\ =14-16\\ =-2\mathrm{m}\end{array}$

Thus, the position of the object at t=2s is -2m.

(c) Determination of the position at t = 3 s

The position of the object at t=3 s is calculated as follows:

$\begin{array}{c}x\left(3\right)=3\left(3\right)-4\left(3^2\right)+\left(3^3\right)\\ =36-36\\ =0\mathrm{m}\end{array}$

Thus, the position of the object at t=3 s is 0m.

(d) Determination of the position at t = 4 s

The position of the object at t=4s is calculated as follows:

$\begin{array}{c}x\left(4\right)=3\left(4\right)-4\left(4^2\right)+\left(4^3\right)\\ =76-64\\ =12\mathrm{m}\end{array}$

Thus, the position of the object at t=4 s is 12m.

(e) Determination of the displacements between the given time interval

The position of the object at t=0 s is calculated as follows:

$\begin{array}{c}x\left(0\right)=3\left(0\right)-4\left(0^2\right)+\left(0^3\right)\\ =0\mathrm{m}\end{array}$

The displacement of the object between t=0s and t=4 s is calculated as follows:

$\begin{array}{c}\Delta x=x\left(4\right)-x\left(0\right)\\ =12-0\\ =+12\mathrm{m}\end{array}$

Thus, the displacement of the object is +12 m.

(f) Determination of the average velocity between the given time interval

The displacement of the object between t=2 s and t=4 s is calculated as follows:

$\begin{array}{c}\Delta x=x\left(4\right)-x\left(2\right)\\ =12-\left(-2\right)\\ =+14\mathrm{m}\end{array}$

The average velocity of the object is calculated as follows:

$\begin{array}{c}{v}_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}\\ =\frac{+14\mathrm{m}}{4\mathrm{s}-2\mathrm{s}}\\ =+7\mathrm{m}/\mathrm{s}\end{array}$

(g) Plot the x-t graph and determination of the average velocity from it

Below is the graph of position vs. time for$0\le t\le 4\sec$ . To find the average velocity between t=2 sec to t=4 sec , you can connect the points on the graph as shown by the black line in the graph. The slope of this black line will be the average velocity.