object d:
graph of velocity - time for object d: velocity (m/s) on y - axis, time (s) on x - axis. from t = 0 to t = 2 s, velocity is - 4 m/s (horizontal line). from t = 2 s to t = 8 s, velocity increases linearly from - 4 m/s to 8 m/s

a. give a written description of the motion.

b. sketch a motion map. be sure to include both velocity and acceleration vectors.
vel:
number line with 0m and + direction, tick marks
acc:

c. determine the displacement from t = 0 s to t = 4 s.

d. determine the displacement from t = 4 s to t = 8 s.

e. determine the displacement from t = 2 s to t = 6 s.

f. determine the object’s acceleration at t = 4 s.

g. sketch a possible x - t graph for the motion of the object.
explain why your graph is only one of many possible graphs.

Question

object d:
graph of velocity - time for object d: velocity (m/s) on y - axis, time (s) on x - axis. from t = 0 to t = 2 s, velocity is - 4 m/s (horizontal line). from t = 2 s to t = 8 s, velocity increases linearly from - 4 m/s to 8 m/s

a. give a written description of the motion.

b. sketch a motion map. be sure to include both velocity and acceleration vectors.
vel:
number line with 0m and + direction, tick marks
acc:

c. determine the displacement from t = 0 s to t = 4 s.

d. determine the displacement from t = 4 s to t = 8 s.

e. determine the displacement from t = 2 s to t = 6 s.

f. determine the object’s acceleration at t = 4 s.

g. sketch a possible x - t graph for the motion of the object.
explain why your graph is only one of many possible graphs.

Accepted Answer

### Part a: Description of Motion
# Brief Explanations:
- From $ t = 0 $ to $ t = 2 $ s: The velocity is constant at $ -4 \, \text{m/s} $, so the object moves with uniform motion (constant velocity) in the negative direction.
- From $ t = 2 $ to $ t = 8 $ s: The velocity - time graph is a straight line with a positive slope, meaning the object is accelerating (constant acceleration) in the positive direction. The velocity changes from $ -4 \, \text{m/s} $ at $ t = 2 $ s to $ 8 \, \text{m/s} $ at $ t = 8 $ s. Initially (from $ t = 2 $ to $ t = 4 $ s), the velocity is negative but increasing (becoming less negative), so the object is slowing down while moving in the negative direction. At $ t = 4 $ s, the velocity becomes zero, and then from $ t = 4 $ to $ t = 8 $ s, the velocity is positive and increasing, so the object is speeding up while moving in the positive direction.
# Answer:
From $ t = 0 $ to $ t = 2 \, \text{s} $, the object moves with a constant velocity of $ - 4\,\text{m/s} $ (uniform motion in the negative direction). From $ t = 2 \, \text{s} $ to $ t = 8 \, \text{s} $, the object has a constant positive acceleration. From $ t = 2 \, \text{s} $ to $ t = 4 \, \text{s} $, it slows down while moving in the negative direction (velocity goes from $ - 4\,\text{m/s} $ to $ 0\,\text{m/s} $). From $ t = 4 \, \text{s} $ to $ t = 8 \, \text{s} $, it speeds up while moving in the positive direction (velocity goes from $ 0\,\text{m/s} $ to $ 8\,\text{m/s} $).

### Part c: Displacement from $ t = 0 $ s to $ t = 4 $ s
# Explanation:
## Step 1: Analyze the velocity - time graph segments
- From $ t = 0 $ to $ t = 2 $ s: The velocity $ v_1=-4\,\text{m/s} $, time interval $ \Delta t_1 = 2 - 0=2\,\text{s} $. The displacement $ d_1=v_1\times\Delta t_1 $ (since velocity is constant).
- From $ t = 2 $ to $ t = 4 $ s: The velocity - time graph is a straight line. The initial velocity at $ t = 2 $ s is $ v_{2i}=-4\,\text{m/s} $, final velocity at $ t = 4 $ s is $ v_{2f} = 0\,\text{m/s} $. The displacement for a uniformly accelerated motion (or motion with changing velocity, we can use the average velocity formula $ d_2=\frac{v_{2i}+v_{2f}}{2}\times\Delta t_2 $, where $ \Delta t_2=4 - 2 = 2\,\text{s} $)
## Step 2: Calculate $ d_1 $
$ d_1=-4\times2=-8\,\text{m} $
## Step 3: Calculate $ d_2 $
$ d_2=\frac{-4 + 0}{2}\times2=-4\,\text{m} $
## Step 4: Total displacement $ d=d_1 + d_2 $
$ d=-8+( - 4)=-12\,\text{m} $
# Answer:
The displacement from $ t = 0 \, \text{s} $ to $ t = 4 \, \text{s} $ is $ - 12\,\text{m} $ (or $ 12\,\text{m} $ in the negative direction).

### Part d: Displacement from $ t = 4 $ s to $ t = 8 $ s
# Explanation:
## Step 1: Analyze the velocity - time graph
From $ t = 4 $ to $ t = 8 $ s, the velocity - time graph is a straight line. The initial velocity $ v_{i}=0\,\text{m/s} $ at $ t = 4 $ s, final velocity $ v_{f}=8\,\text{m/s} $ at $ t = 8 $ s, and time interval $ \Delta t=8 - 4 = 4\,\text{s} $. We can use the average velocity formula $ d=\frac{v_{i}+v_{f}}{2}\times\Delta t $ (since the motion is uniformly accelerated).
## Step 2: Calculate the displacement
$ d=\frac{0 + 8}{2}\times4=4\times4 = 16\,\text{m} $
# Answer:
The displacement from $ t = 4 \, \text{s} $ to $ t = 8 \, \text{s} $ is $ 16\,\text{m} $ (in the positive direction).

### Part e: Displacement from $ t = 2 $ s to $ t = 6 $ s
# Explanation:
## Step 1: Divide the time interval
- From $ t = 2 $ to $ t = 4 $ s: Initial velocity $ v_{1i}=-4\,\text{m/s} $, final velocity $ v_{1f}=0\,\text{m/s} $, $ \Delta t_1=4 - 2 = 2\,\text{s} $. Displacement $ d_1=\frac{v_{1i}+v_{1f}}{2}\times\Delta t_1 $
- From $ t = 4 $ to $ t = 6 $ s: Initial velocity $ v_{2i}=0\,\text{m/s} $, final velocity $ v_{2f}=4\,\text{m/s} $ (since at $ t = 6 $ s, from the graph, velocity is $ 4\,\text{m/s} $), $ \Delta t_2=6 - 4 = 2\,\text{s} $. Displacement $ d_2=\frac{v_{2i}+v_{2f}}{2}\times\Delta t_2 $
## Step 2: Calculate $ d_1 $
$ d_1=\frac{-4 + 0}{2}\times2=-4\,\text{m} $
## Step 3: Calculate $ d_2 $
$ d_2=\frac{0 + 4}{2}\times2 = 4\,\text{m} $
## Step 4: Total displacement $ d=d_1 + d_2 $
$ d=-4 + 4=0\,\text{m} $
# Answer:
The displacement from $ t = 2 \, \text{s} $ to $ t = 6 \, \text{s} $ is $ 0\,\text{m} $.

### Part f: Acceleration at $ t = 4 $ s
# Explanation:
## Step 1: Recall the formula for acceleration
Acceleration $ a=\frac{\Delta v}{\Delta t} $. We can use the slope of the velocity - time graph. From $ t = 2 $ s to $ t = 8 $ s, the velocity changes from $ v_1=-4\,\text{m/s} $ (at $ t = 2 $ s) to $ v_2 = 8\,\text{m/s} $ (at $ t = 8 $ s), and the time interval $ \Delta t=8 - 2=6\,\text{s} $
## Step 2: Calculate acceleration
$ a=\frac{8-( - 4)}{6}=\frac{12}{6}=2\,\text{m/s}^2 $
Since the acceleration is constant from $ t = 2 $ s to $ t = 8 $ s, the acceleration at $ t = 4 $ s is also $ 2\,\text{m/s}^2 $
# Answer:
The acceleration at $ t = 4 \, \text{s} $ is $ 2\,\text{m/s}^2 $.

### Part g: Sketch of $ x - t $ graph and Explanation
# Brief Explanations:
- **Sketch of $ x - t $ graph**:
    - From $ t = 0 $ to $ t = 2 $ s: The object has a constant velocity of $ - 4\,\text{m/s} $. So the $ x - t $ graph is a straight line with a slope of $ - 4\,\text{m/s} $. Let's assume the initial position $ x_0 = 0\,\text{m} $ (we can assume any initial position, which is why there are many possible graphs). At $ t = 2 $ s, $ x_2=0+( - 4)\times2=-8\,\text{m} $
    - From $ t = 2 $ to $ t = 8 $ s: The object has a constant acceleration of $ 2\,\text{m/s}^2 $. The equation of motion is $ x=x_2+v_{2}(t - 2)+\frac{1}{2}a(t - 2)^2 $, where $ v_{2}=-4\,\text{m/s} $ (velocity at $ t = 2 $ s) and $ a = 2\,\text{m/s}^2 $. The $ x - t $ graph will be a parabola opening upwards (since the acceleration is positive) with a vertex - like behavior. At $ t = 4 $ s, $ v = 0$, so the slope of the $ x - t $ graph (which is velocity) is zero at $ t = 4 $ s.
- **Explanation of multiple possible graphs**:
The $ x - t $ graph depends on the initial position of the object. We assumed an initial position of $ x = 0\,\text{m} $ at $ t = 0 $ s, but the object could have started from any position (e.g., $ x = 5\,\text{m} $ or $ x=-10\,\text{m} $ at $ t = 0 $ s). As long as the velocity - time behavior (derived from the slope of the $ x - t $ graph) matches the given $ v - t $ graph, the $ x - t $ graph is valid. So there are infinitely many possible $ x - t $ graphs, each corresponding to a different initial position.
# Answer:
- **Sketch**:
    - From $ t = 0 $ to $ t = 2 \, \text{s} $: A straight line with slope $ - 4\,\text{m/s} $ (e.g., from $ (0,0) $ to $ (2, - 8) $ if initial position is $ 0\,\text{m} $).
    - From $ t = 2 \, \text{s} $ to $ t = 8 \, \text{s} $: A parabola opening upwards. At $ t = 4 \, \text{s} $, the slope of the parabola (velocity) is $ 0\,\text{m/s} $, at $ t = 8 \, \text{s} $, the slope (velocity) is $ 8\,\text{m/s} $.
- **Explanation**: The $ x - t $ graph depends on the initial position of the object. Different initial positions (e.g., starting at $ x = 5\,\text{m} $ or $ x=-10\,\text{m} $ at $ t = 0 $) will result in different $ x - t $ graphs, as long as the velocity - time relationship (from the slope of $ x - t $) matches the given $ v - t $ graph. So our graph is just one of many possible graphs.

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QUESTION IMAGE

Question

Explanation:

Part a: Description of Motion

Step 1: Analyze the velocity - time graph segments

Step 2: Calculate \( d_1 \)

Step 3: Calculate \( d_2 \)

Step 4: Total displacement \( d=d_1 + d_2 \)

Step 1: Analyze the velocity - time graph

Step 2: Calculate the displacement

Answer:

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