Kinematics Motion Graphs Worksheet - Studocu - Free Printable

Let's go through each question on the worksheet step by step, solving and explaining each one.

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Question 1:

A cart travels with a constant nonzero acceleration along a straight line. Which graph best represents the relationship between the distance the cart travels and time of travel?

Answer: Graph (1)

Explanation:
When an object moves with constant nonzero acceleration, its motion follows the kinematic equation:

$$
d = v_0 t + \frac{1}{2} a t^2
$$

If it starts from rest ($v_0 = 0$), then:
$$
d = \frac{1}{2} a t^2
$$

This is a quadratic relationship between distance and time — so the graph of distance vs. time should be a parabola opening upward.

- Graph (1): Parabolic curve → correct.
- Graph (2): Linear decrease → not possible for positive acceleration.
- Graph (3): Arc-shaped (peaks and returns) → implies deceleration and reversal, not constant acceleration.
- Graph (4): Straight line → constant speed, not accelerating.

✔ Correct answer: (1)

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Questions 2–4: Car Motion Problem

Given:
- A car starts from rest: $v_0 = 0$
- Accelerates at $a = 1.0\ \text{m/s}^2$ for $t = 10\ \text{s}$
- Then moves at constant speed for another 20 seconds.

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Question 2: Determine the speed of the car at the end of the first 10 seconds.

Use the kinematic equation:
$$
v = v_0 + a t
$$

Substitute:
$$
v = 0 + (1.0\ \text{m/s}^2)(10\ \text{s}) = 10\ \text{m/s}
$$

✔ Answer: 10 m/s

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Question 3: Construct a graph of speed vs. time for the entire 30-second interval.

We have two phases:

1. First 10 seconds: Constant acceleration → speed increases linearly from 0 to 10 m/s.
- Plot a straight line from (0, 0) to (10, 10).

2. Next 20 seconds (10 to 30 s): Constant speed → speed remains at 10 m/s.
- Draw a horizontal line from (10, 10) to (30, 10).

Graph Description:
- X-axis: Time (0 to 30 s)
- Y-axis: Speed (0 to 20 m/s)
- Line goes up diagonally from (0,0) to (10,10), then flat to (30,10)

✔ You would draw this on the grid provided.

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Question 4: Calculate the distance the car travels in the first 10 seconds.

Use the kinematic equation:
$$
d = v_0 t + \frac{1}{2} a t^2
$$

Plug in values:
$$
d = (0)(10) + \frac{1}{2}(1.0\ \text{m/s}^2)(10\ \text{s})^2 = 0 + 0.5 \times 1.0 \times 100 = 50\ \text{m}
$$

Alternatively, use average velocity:
$$
v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{0 + 10}{2} = 5\ \text{m/s}
$$
$$
d = v_{\text{avg}} \times t = 5\ \text{m/s} \times 10\ \text{s} = 50\ \text{m}
$$

✔ Answer: 50 meters

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Question 5: A student throws a baseball vertically upward and catches it. Which graph best represents velocity vs. time? (Upward is positive.)

Answer: Graph (4)

Explanation:
- The ball starts with positive velocity (thrown up).
- It slows down due to gravity (acceleration = $-g$).
- At the top, velocity = 0.
- Then it falls down, gaining negative velocity.
- Velocity changes linearly with time because acceleration is constant.

So, the velocity-time graph is a straight line starting at positive value, decreasing linearly through zero, to negative values.

- Graph (1): Symmetric triangle, but doesn’t go negative → incorrect.
- Graph (2): Curved, not linear → wrong (velocity change is uniform).
- Graph (3): Velocity becomes positive again → implies reverse throw → no.
- Graph (4): Starts high, decreases linearly, crosses zero, continues downward → correct.

✔ Answer: (4)

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Question 6: Displacement vs. Time Graph – What was the total distance traveled during 10 seconds?

Given: Displacement vs. time graph.

Let’s analyze the graph:

- From $t = 0$ to $t = 4$: displacement increases from 0 to 8 m → moving forward.
- From $t = 4$ to $t = 6$: displacement stays at 8 m → at rest.
- From $t = 6$ to $t = 9$: displacement increases to 16 m → moving forward again.
- From $t = 9$ to $t = 10$: displacement decreases back to 8 m → moving backward.

Now calculate total distance traveled (not displacement):

- $0 \to 4\ \text{s}$: moved from 0 to 8 m → distance = 8 m
- $4 \to 6\ \text{s}$: no movement → 0 m
- $6 \to 9\ \text{s}$: moved from 8 m to 16 m → distance = 8 m
- $9 \to 10\ \text{s}$: moved from 16 m to 8 m → distance = 8 m

Total distance = $8 + 0 + 8 + 8 = 24\ \text{m}$

✔ Answer: 4. 24 m

Note: Displacement is final minus initial: $8\ \text{m} - 0 = 8\ \text{m}$, but distance is total path length = 24 m.

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✔ Final Answers Summary:

1. (1)
2. 10 m/s
3. Speed vs. Time graph: Up to (10,10), then flat to (30,10)
4. 50 m
5. (4)
6. (4) 24 m

Let me know if you'd like a visual sketch or further explanation!