Worksheet for analyzing kinematic graphs, showing position vs. time graphs and spaces to sketch velocity and acceleration vs. time graphs.

Let's go through each of the position vs. time (x-t) graphs and determine the corresponding velocity vs. time (v-t) and acceleration vs. time (a-t) graphs based on the principles of kinematics.

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🔷 Key Concepts:

1. Slope of position vs. time graph = velocity
2. Slope of velocity vs. time graph = acceleration
3. Area under velocity vs. time graph = change in position
4. Constant slope → constant velocity → zero acceleration
5. Curved line → changing velocity → non-zero acceleration

We'll analyze each case step by step.

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✔ Problem 1:

Position vs. Time: Horizontal line (constant position)

- Velocity: Slope = 0 → constant velocity = 0
- Acceleration: Since velocity is constant (zero), acceleration = 0

✔ Sketch:
- v-t: Flat line at v = 0
- a-t: Flat line at a = 0

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✔ Problem 2:

Position vs. Time: Straight line with positive slope

- Velocity: Constant positive slope → constant positive velocity
- Acceleration: Velocity is constant → acceleration = 0

✔ Sketch:
- v-t: Horizontal line above the axis (positive)
- a-t: Flat line at a = 0

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✔ Problem 3:

Position vs. Time: Straight line with negative slope

- Velocity: Constant negative slope → constant negative velocity
- Acceleration: Velocity is constant → acceleration = 0

✔ Sketch:
- v-t: Horizontal line below the axis (negative)
- a-t: Flat line at a = 0

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✔ Problem 4:

Position vs. Time: Upward-opening parabola (curving upward)

- This is quadratic → indicates constant positive acceleration
- The slope increases over time → velocity increases linearly
- So, velocity is increasing linearly, meaning acceleration is constant and positive

✔ Sketch:
- v-t: Straight line starting from some initial velocity (possibly zero) and increasing positively
- a-t: Horizontal line above axis (constant positive acceleration)

> Note: If the curve starts at origin and is symmetric, initial velocity may be zero.

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✔ Problem 5:

Position vs. Time: Downward-curving arc (like a falling object)

- Shape is downward concave → decreasing slope → velocity decreasing
- But since it's curving downward, slope becomes more negative → velocity is becoming more negative (if going down)
- This implies negative acceleration (e.g., gravity pulling down)
- However, this is a parabolic shape opening downward → constant negative acceleration

✔ Sketch:
- v-t: Linear line with negative slope (decreasing velocity)
- a-t: Horizontal line below axis (constant negative acceleration)

> Example: Object thrown upward and falling back — but here position is decreasing → could be free fall.

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✔ Problem 6:

Position vs. Time: Upward-curving arc (concave up)

- Similar to #4, but now increasing then leveling off? Wait — actually, it looks like concave up, so slope increasing → velocity increasing
- But wait: It starts at origin, curves upward, then flattens? Or does it keep increasing?

Actually, the curve is concave up, so the slope is increasing → velocity increasing → positive acceleration

But if it’s a parabola (as implied), then constant acceleration

✔ So:
- v-t: Line with positive slope (increasing velocity)
- a-t: Constant positive value

> Note: Initial velocity might be zero or positive.

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✔ Problem 7:

Position vs. Time: Curve that starts high and decreases rapidly, curving upward (convex up)

- This is a downward parabola → same as problem 5?
- Actually, it's concave up, so slope is increasing (becoming less negative or turning positive?)

Wait: Position is decreasing, but the curve is bending upward → slope is increasing (less steep negative)

So:
- Velocity is negative (moving left/down), but getting less negative → slowing down in the negative direction → positive acceleration
- And if it's a parabola, acceleration is constant

✔ So:
- v-t: Starts at negative velocity, increases toward zero (linear)
- a-t: Constant positive value

> Example: Object moving left with decreasing speed → decelerating in negative direction → positive acceleration

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✔ Problem 8:

Position vs. Time: Parabola opening downward (symmetric arch)

- This is a classic projectile motion type: object goes up and comes down
- Symmetric about vertex
- At peak, slope = 0 → velocity = 0
- Before peak: slope positive → velocity positive
- After peak: slope negative → velocity negative
- Acceleration is constant (gravity): always negative (downward)

✔ So:
- v-t: Straight line with negative slope, starting positive, going through zero, then negative
- a-t: Constant negative value (horizontal line below axis)

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## 📊 Summary Table:

| Problem | x-t Graph | v-t Graph | a-t Graph |
|--------|-----------|-----------|-----------|
| 1 | Flat line | Flat at 0 | Flat at 0 |
| 2 | Positive slope line | Flat positive | Flat at 0 |
| 3 | Negative slope line | Flat negative | Flat at 0 |
| 4 | Upward parabola | Increasing line (positive slope) | Flat positive |
| 5 | Downward parabola | Decreasing line (negative slope) | Flat negative |
| 6 | Upward parabola | Increasing line (positive slope) | Flat positive |
| 7 | Downward convex (concave up) | Increasing from negative to zero | Flat positive |
| 8 | Symmetric arch (up then down) | Line decreasing from positive to negative | Flat negative |

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## ✔ Final Instructions for Drawing:

For each stack:

Step-by-step guide:

1. Look at the slope of x-t → draw v-t accordingly:
- Constant slope → straight horizontal v-t
- Increasing slope → rising v-t line
- Decreasing slope → falling v-t line

2. Look at how slope changes → draw a-t:
- Constant slope → zero acceleration (flat a-t)
- Changing slope → non-zero acceleration
- If slope increases uniformly → constant acceleration
- If slope changes at constant rate → straight line in v-t → flat a-t

3. Draw a-t:
- If v-t is flat → a-t is zero
- If v-t is straight line → a-t is constant (non-zero)
- If v-t is curved → a-t is changing

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✔ Answers (Sketching Guide):

#### Problem 1
- v-t: Horizontal line at 0
- a-t: Horizontal line at 0

#### Problem 2
- v-t: Horizontal line above axis
- a-t: Horizontal line at 0

#### Problem 3
- v-t: Horizontal line below axis
- a-t: Horizontal line at 0

#### Problem 4
- v-t: Diagonal line upward (starting from origin or slightly above)
- a-t: Horizontal line above axis

#### Problem 5
- v-t: Diagonal line downward (starting from positive or zero)
- a-t: Horizontal line below axis

#### Problem 6
- v-t: Diagonal line upward (starting from origin)
- a-t: Horizontal line above axis

#### Problem 7
- v-t: Diagonal line upward from negative velocity toward zero
- a-t: Horizontal line above axis

#### Problem 8
- v-t: Diagonal line downward from positive to negative velocity (crosses zero)
- a-t: Horizontal line below axis

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## 💡 Tips:
- Use grid lines to match scales.
- Label axes clearly.
- For curved x-t graphs, assume uniform acceleration (since it's Unit 3: Uniform Acceleration).
- Only straight lines in v-t mean constant acceleration; only flat lines in a-t mean uniform acceleration.

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Let me know if you'd like a visual sketch description or help drawing them!