object c:
a. give a written description of the motion.
b. sketch a motion map. be sure to include both velocity and acceleration vectors.
vel:
acc:
c. determine the displacement from t = 0s to t = 4 s.
d. determine the displacement from t = 4 s to t = 8 s.
e. determine the average acceleration of the object’s motion.
f. sketch a possible x-t graph for the motion of the object.
explain why your graph is only one of many possible graphs.

Question

Accepted Answer

### Part a: Written Description of Motion
# Brief Explanations:
The velocity - time (v - t) graph shows that the object starts with an initial velocity of $v_0=- 3\space m/s$ (negative sign indicates motion in the negative direction) at $t = 0\space s$. The velocity is decreasing linearly (since the v - t graph is a straight line with negative slope), which means the object is accelerating in the negative direction (decelerating in the positive direction or accelerating negatively). So the object is moving in the negative direction and speeding up (because acceleration and velocity are in the same direction, both negative) as time progresses.
# Answer:
The object starts with an initial velocity of $-3\space m/s$ (moving in the negative direction) at $t = 0\space s$. It has a constant negative acceleration (since the $v - t$ graph is a straight line with negative slope), so it is moving in the negative direction and speeding up as time increases.

### Part c: Displacement from $t = 0\space s$ to $t=4\space s$
# Explanation:
## Step 1: Recall the formula for displacement from a $v - t$ graph
The displacement of an object from a velocity - time graph is given by the area under the $v - t$ curve between the two time instants. For a linear $v - t$ graph (a straight line), the area under the curve is the area of a trapezoid (or a triangle in some cases, here it's a trapezoid with bases $v_0$ and $v_4$ and height $\Delta t$). The formula for the area of a trapezoid is $A=\frac{1}{2}(b_1 + b_2)h$, where $b_1$ and $b_2$ are the lengths of the two parallel sides and $h$ is the distance between them. In the context of $v - t$ graph, $b_1=v(0)$, $b_2 = v(4)$ and $h=\Delta t=t_2 - t_1$

First, we need to find $v(4)$. The slope of the $v - t$ graph is constant (since it's a straight line). The slope $m=\frac{\Delta v}{\Delta t}$. From $t = 0$ to $t = 8$, $v(0)=-3\space m/s$ and $v(8)=-9\space m/s$. So the slope $m=\frac{-9-(-3)}{8 - 0}=\frac{-6}{8}=- 0.75\space m/s^2$

At $t = 4\space s$, $v(4)=v(0)+m\times(4 - 0)=-3+(-0.75)\times4=-3 - 3=-6\space m/s$

## Step 2: Calculate the area (displacement)
The time interval $\Delta t=4 - 0 = 4\space s$. The two velocities at $t = 0$ and $t = 4$ are $v_0=-3\space m/s$ and $v_4=-6\space m/s$

Using the formula for the area of a trapezoid $A=\frac{1}{2}(v_0 + v_4)\times\Delta t$

Substitute $v_0=-3\space m/s$, $v_4=-6\space m/s$ and $\Delta t = 4\space s$ into the formula:

$A=\frac{1}{2}(-3+(-6))\times4=\frac{1}{2}(-9)\times4=- 18\space m$

# Answer:
The displacement from $t = 0\space s$ to $t = 4\space s$ is $-18\space m$ (the negative sign indicates displacement in the negative direction)

### Part d: Displacement from $t = 4\space s$ to $t = 8\space s$
# Explanation:
## Step 1: Recall the formula for displacement from $v - t$ graph
Again, displacement is the area under the $v - t$ curve. From $t = 4$ to $t = 8$, the $v - t$ graph is a straight line. We can use the formula for the area of a trapezoid (or a triangle, here it's a trapezoid with bases $v(4)$ and $v(8)$ and height $\Delta t$)

We know $v(4)=-6\space m/s$ and $v(8)=-9\space m/s$, $\Delta t=8 - 4 = 4\space s$

## Step 2: Calculate the area (displacement)
Using the trapezoid area formula $A=\frac{1}{2}(b_1 + b_2)h$, where $b_1 = v(4)=-6\space m/s$, $b_2=v(8)=-9\space m/s$ and $h = 4\space s$

$A=\frac{1}{2}(-6+(-9))\times4=\frac{1}{2}(-15)\times4=-30\space m$

# Answer:
The displacement from $t = 4\space s$ to $t = 8\space s$ is $-30\space m$ (negative sign indicates displacement in the negative direction)

### Part e: Average Acceleration from $t = 0\space s$ to $t = 8\space s$
# Explanation:
## Step 1: Recall the formula for average acceleration
The average acceleration $a_{avg}=\frac{\Delta v}{\Delta t}$, where $\Delta v=v_f - v_i$ and $\Delta t=t_f - t_i$

## Step 2: Identify $v_i$, $v_f$, $t_i$ and $t_f$
Here, $v_i=v(0)=-3\space m/s$, $v_f=v(8)=-9\space m/s$, $t_i = 0\space s$ and $t_f=8\space s$

## Step 3: Calculate $\Delta v$ and $\Delta t$
$\Delta v=-9-(-3)=-6\space m/s$

$\Delta t=8 - 0 = 8\space s$

## Step 4: Calculate average acceleration
$a_{avg}=\frac{\Delta v}{\Delta t}=\frac{-6}{8}=- 0.75\space m/s^2$

# Answer:
The average acceleration of the object's motion is $-0.75\space m/s^2$

### Part f: Sketch of $x - t$ graph and Explanation
# Brief Explanations:
#### Sketch of $x - t$ graph:
- The object has a negative velocity and negative acceleration, so the $x - t$ graph will be a curve (since acceleration is non - zero, the graph is not linear) that is concave down (because acceleration is negative, the second derivative of $x$ with respect to $t$ (which is acceleration) is negative). At $t = 0$, let's assume the initial position $x(0) = 0$ (we can assume any initial position, which is why there are multiple possible graphs). The slope of the $x - t$ graph at $t = 0$ is equal to the initial velocity $v(0)=-3\space m/s$ (negative slope). As time increases, the slope of the $x - t$ graph becomes more negative (since the object is speeding up in the negative direction), so the curve will be a downward - opening parabola - like curve (because acceleration is constant and negative, the position function $x(t)=x_0+v_0t+\frac{1}{2}at^2$, with $x_0$ as initial position, $v_0=-3\space m/s$ and $a=-0.75\space m/s^2$)

#### Explanation why it's one of many possible graphs:
The $x - t$ graph depends on the initial position of the object. We assumed the initial position $x(0) = 0$, but the initial position could be any real number. For example, if the initial position was $x(0)=5\space m$, the entire $x - t$ graph would be shifted up by $5\space m$ while still having the same shape (concave down, with increasing negative slope). Also, the scale of the axes (the units and the range of $x$ and $t$ we choose to plot) can vary, but more importantly, the initial position is arbitrary. So there are infinitely many possible $x - t$ graphs, each corresponding to a different initial position of the object.
# Answer:
- **Sketch of $x - t$ graph**: Assume initial position $x(0) = 0$. The graph is a concave - down curve (since acceleration $a<0$). At $t = 0$, the slope is $-3\space m/s$ (negative). As $t$ increases, the slope becomes more negative (e.g., at $t = 4$, slope is $-6\space m/s$; at $t = 8$, slope is $-9\space m/s$). The equation of the position function is $x(t)=x_0+v_0t+\frac{1}{2}at^2$, with $x_0 = 0$, $v_0=-3\space m/s$ and $a=-0.75\space m/s^2$, so $x(t)=-3t - 0.375t^2$ (a downward - opening parabola).
- **Explanation**: The $x - t$ graph is only one of many possible graphs because the initial position $x(0)$ of the object is not specified. The initial position can be any real number, and each different initial position will result in a vertically shifted version of the same - shaped $x - t$ graph. Also, the coordinate system (scale of $x$ and $t$ axes) can be chosen differently, but the main reason is the arbitrary nature of the initial position.

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QUESTION IMAGE

Question

Explanation:

Part a: Written Description of Motion

Step 1: Recall the formula for displacement from a \(v - t\) graph

Step 2: Calculate the area (displacement)

Step 1: Recall the formula for displacement from \(v - t\) graph

Step 2: Calculate the area (displacement)

Answer:

Part c: Displacement from \(t = 0\space s\) to \(t=4\space s\)