Physics worksheet for analyzing motion using position-time and velocity-time graphs.

Final Answer:
a) Graphs should show:
- Position-Time graph: points at (0,0), (1,2), (2,4), (3,6), (4,6), (5,4), (6,2), (7,0), (8,−2); connect with straight line segments.
- Velocity-Time graph: horizontal segments at v = +2 from t = 0 to 4 s, v = 0 from t = 4 to 5 s, v = −2 from t = 5 to 8 s.

b) Displacement from 2 s to 5 s = position at 5 s − position at 2 s = 4 − 4 = 0 m.

c) From 0–4 s: displacement = 6 − 0 = +6 m
 From 5–8 s: displacement = −2 − 4 = −6 m

d) Total displacement = final position − initial position = −2 − 0 = −2 m
Total distance = sum of absolute segment lengths:
|2−0| + |4−2| + |6−4| + |6−6| + |4−6| + |2−4| + |0−2| + |−2−0| = 2+2+2+0+2+2+2+2 = 14 m

e) The area under the velocity-time graph gives displacement (with sign).
Positive area (above axis) = +8 m (from 0–4 s: 4 s × 2 m/s)
Zero area (4–5 s) = 0
Negative area (5–8 s) = −6 m (3 s × −2 m/s)
Total area = +8 + 0 − 6 = +2? Wait — correction:
Actually, from table:
v = +2 from t=0 to t=4 → area = 4 × 2 = +8
v = 0 from t=4 to t=5 → area = 0
v = −2 from t=5 to t=8 → area = 3 × (−2) = −6
Total area = +2 — but displacement is −2. Contradiction? Let’s recheck positions:

Given table:
t: 0 1 2 3 4 5 6 7 8
x: 0 2 4 6 6 4 2 0 −2

So from t=0 to 4: x goes 0→6 → Δx = +6
But velocity = Δx/Δt = (6−0)/4 = +1.5? Wait — no: between each 1-second step:
0→1: +2 → v = +2
1→2: +2 → v = +2
2→3: +2 → v = +2
3→4: 0 → v = 0
Ah! Mistake in earlier assumption.

Let’s compute velocity from position changes (Δx/Δt per 1 s interval):
t=0→1: (2−0)/1 = +2
t=1→2: (4−2)/1 = +2
t=2→3: (6−4)/1 = +2
t=3→4: (6−6)/1 = 0
t=4→5: (4−6)/1 = −2
t=5→6: (2−4)/1 = −2
t=6→7: (0−2)/1 = −2
t=7→8: (−2−0)/1 = −2

So velocity is:
+2 for t ∈ [0,3] (3 seconds),
0 for t ∈ [3,4],
−2 for t ∈ [4,8] (4 seconds)

Thus velocity-time graph:
- v = +2 from t=0 to t=3
- v = 0 from t=3 to t=4
- v = −2 from t=4 to t=8

Now recalculate:

b) Displacement from 2 s to 5 s = x(5) − x(2) = 4 − 4 = 0 m ✔

c) 0–4 s: x(4) − x(0) = 6 − 0 = +6 m
 5–8 s: x(8) − x(5) = −2 − 4 = −6 m ✔

d) Total displacement = x(8) − x(0) = −2 − 0 = −2 m
Total distance = |2|+|2|+|2|+|0|+|−2|+|−2|+|−2|+|−2| = 2+2+2+0+2+2+2+2 = 14 m ✔

e) Area under v-t graph:
+2 × 3 s = +6
0 × 1 s = 0
−2 × 4 s = −8
Total area = +6 + 0 − 8 = −2 m → matches displacement.

So final corrected answers:

b) 0 m
c) +6 m and −6 m
d) displacement = −2 m, distance = 14 m
e) The signed area under the velocity-time graph equals displacement; total area (ignoring sign) gives distance only if motion is unidirectional—but here we must sum absolute areas for distance. In this case, signed area = −2 m = displacement.

Since the question asks for *only* the final answer (and ExplainAnswer etc. are false), we give concise numeric answers as expected in worksheet blanks:

Final Answer:
b) 0 m
c) 0–4 s: +6 m; 5–8 s: −6 m
d) displacement = −2 m, distance = 14 m
e) The displacement equals the net (signed) area under the velocity-time graph; here, +6 m + (−8 m) = −2 m.