At time `t=0 s` , an object is observed at `x=0` m , and its position along the `x` axis follows this expression : `x= -3t +t^(3)`, where the units for distance and time are meter and second, respectively. What is the object's displacement `Deltax` between `t=1.0` s and `t=3.0` s ?

Options

A. `+20 m`
B. `+10`m
C. `-20m`
D. `+2` m

Answer

To find the object's displacement \(\Delta x\) between \(t = 1.0\) s and \(t = 3.0\) s, we will follow these steps:

### Step 1: Calculate the position at \(t = 1.0\) s
We use the position equation given:
\[ x(t) = -3t + t^3 \]
Substituting \(t = 1.0\) s:
\[ x(1) = -3(1) + (1)^3 \]
Calculating this:
\[ x(1) = -3 + 1 = -2 \text{ m} \]

### Step 2: Calculate the position at \(t = 3.0\) s
Now, we substitute \(t = 3.0\) s into the same position equation:
\[ x(3) = -3(3) + (3)^3 \]
Calculating this:
\[ x(3) = -9 + 27 = 18 \text{ m} \]

### Step 3: Calculate the displacement \(\Delta x\)
Displacement is defined as the change in position:
\[ \Delta x = x(t_2) - x(t_1) \]
Where \(t_2 = 3.0\) s and \(t_1 = 1.0\) s. Thus:
\[ \Delta x = x(3) - x(1) = 18 - (-2) \]
Calculating this:
\[ \Delta x = 18 + 2 = 20 \text{ m} \]

### Final Answer
The object's displacement \(\Delta x\) between \(t = 1.0\) s and \(t = 3.0\) s is \(20\) meters.

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