Graphing Motion Kinematic WS Solutions - Free Printable

Let's carefully analyze the displacement-time graph and answer each question step by step. The graph shows displacement (in meters) on the vertical axis and time (in seconds) on the horizontal axis.

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Key Concept:

- Displacement is the change in position from a starting point.
- On a displacement-time graph, the slope of the line represents velocity.
- A flat (horizontal) line means the object is at rest (velocity = 0).
- A positive slope means motion in the positive direction.
- A negative slope means motion in the negative direction.
- Distance traveled is the total path length, but since this is a displacement graph, we must be careful: if the object moves back and forth, distance may differ from displacement.

But note: The questions ask "how far does the object travel", which refers to distance traveled, not displacement.

However, in this graph, there are no loops or back-and-forth motions—the object only changes direction once at t = 20 s.

So we can calculate distance traveled as the absolute value of displacement changes over intervals.

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Let’s go through each question:

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11. How far does the object travel during the first 5 seconds (1 to 5 s)?

From t = 0 to t = 5 s, the displacement goes from 0 m to 5 m.

Even though it says "1 to 5 s", assuming it's a typo and meant to be 0 to 5 s, because the graph starts at t=0.

- Displacement change: $ 5\,\text{m} - 0\,\text{m} = 5\,\text{m} $
- Since the object is moving forward continuously, distance = displacement = 5 m

✔ Answer: 5 m

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12. How far does the object travel during the second 5 seconds (5 to 10 s)?

From t = 5 s to t = 10 s:
- At t = 5 s: displacement = 5 m
- At t = 10 s: displacement = 5 m

So the object stays at 5 m — no change in displacement.

➡️ Object is standing still.

✔ Answer: 0 m

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13. How far does the object travel during the third 5 seconds (10 to 15 s)?

From t = 10 s to t = 15 s:
- At t = 10 s: 5 m
- At t = 15 s: 15 m

Change in displacement: $ 15\,\text{m} - 5\,\text{m} = 10\,\text{m} $

The object moves steadily forward → distance = displacement.

✔ Answer: 10 m

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14. How far does the object travel during the fourth 5 seconds (15 to 20 s)?

From t = 15 s to t = 20 s:
- At t = 15 s: 15 m
- At t = 20 s: 15 m

No change in displacement → object is stationary

✔ Answer: 0 m

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15. How far does the object travel during the last 10 seconds (20 to 30 s)?

From t = 20 s to t = 30 s:
- At t = 20 s: 15 m
- At t = 30 s: 0 m

Displacement change: $ 0 - 15 = -15\,\text{m} $

But distance traveled is the absolute value of the path taken.

Since the object moves straight from 15 m to 0 m (no reversal), the distance is:

$$
| -15 | = 15\,\text{m}
$$

✔ Answer: 15 m

(Note: Your handwritten answer says "-15m", but distance cannot be negative. So correct answer is 15 m.)

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16. During which time interval(s) is the object standing still?

Look for horizontal lines (zero slope):

- From t = 5 s to t = 10 s: displacement is constant at 5 m → stationary
- From t = 15 s to t = 20 s: displacement is constant at 15 m → stationary

Also, what about t = 0 to 5 s? No, it's increasing — moving.

Wait! What about before t = 0? Not relevant.

But wait — your handwritten answer says (0–5s), (15s–20s) — that’s incorrect.

At 0–5s, the object is moving from 0 to 5 m — not standing still.

Only when the graph is flat.

So correct intervals:

- 5 s to 10 s
- 15 s to 20 s

✔ Answer: (5 s to 10 s), (15 s to 20 s)

> ⚠️ You wrote (0–5s), which is wrong — the object is moving then.

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17. Does the object ever change direction? If so, when?

Yes — look at the slope:

- From t = 0 to 5 s: positive slope (moving forward)
- t = 5 to 10 s: flat → stopped
- t = 10 to 15 s: positive slope → moving forward again
- t = 15 to 20 s: flat → stopped
- t = 20 to 30 s: negative slope → moving backward

So the object changes direction at t = 20 s, when it starts moving from 15 m toward 0 m.

✔ Answer: Yes, at t = 20 s

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18. Draw the velocity-time graph for the above scenario

We need to compute velocity in each segment.

Velocity = slope of displacement-time graph = $ \frac{\Delta d}{\Delta t} $

#### Segment A: 0 to 5 s
- $ \Delta d = 5 - 0 = 5\,\text{m} $
- $ \Delta t = 5\,\text{s} $
- $ v = \frac{5}{5} = 1\,\text{m/s} $

→ Constant velocity: +1 m/s

#### Segment B: 5 to 10 s
- $ \Delta d = 0 $
- $ v = 0\,\text{m/s} $

#### Segment C: 10 to 15 s
- $ \Delta d = 15 - 5 = 10\,\text{m} $
- $ \Delta t = 5\,\text{s} $
- $ v = \frac{10}{5} = 2\,\text{m/s} $

→ +2 m/s

#### Segment D: 15 to 20 s
- $ \Delta d = 0 $
- $ v = 0\,\text{m/s} $

#### Segment E: 20 to 30 s
- $ \Delta d = 0 - 15 = -15\,\text{m} $
- $ \Delta t = 10\,\text{s} $
- $ v = \frac{-15}{10} = -1.5\,\text{m/s} $

→ Constant velocity: -1.5 m/s

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Velocity-Time Graph Description:

- From 0 to 5 s: horizontal line at v = +1 m/s
- From 5 to 10 s: horizontal line at v = 0 m/s
- From 10 to 15 s: horizontal line at v = +2 m/s
- From 15 to 20 s: horizontal line at v = 0 m/s
- From 20 to 30 s: horizontal line at v = -1.5 m/s

This would be a step graph with flat segments at different velocities.

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Summary of Answers:

| Question | Answer |
|--------|--------|
| 11 | 5 m |
| 12 | 0 m |
| 13 | 10 m |
| 14 | 0 m |
| 15 | 15 m |
| 16 | (5 s to 10 s), (15 s to 20 s) |
| 17 | Yes, at t = 20 s |
| 18 | Velocity-time graph: steps at +1, 0, +2, 0, -1.5 m/s |

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✔ Corrections made:
- Q11: Corrected assumption (0–5s, not 1–5s)
- Q15: Distance is 15 m, not -15 m
- Q16: Object is not still from 0–5s — it’s moving

Let me know if you'd like a sketch of the velocity-time graph!