IGCSE Physics 0625 · Section 1.2 · Motion — Numerical Solutions
Motion — Numerical Practice: Solutions
A. Speed and Average Speed
A1
A car travels 150 km in 2.5 hours. Calculate its average speed.
1
average speed = total distance / total time
150 ÷ 2.5
✓60 km/h
A2
A runner covers 200 m in 25 s. Calculate her speed.
✓8 m/s
A3
A train travels at 120 km/h for 3 h, stops 30 min, then travels 60 km in 1 h. Find average speed for the whole journey.
1
Total distance
(120 × 3) + 60 = 420 km
2
Total time (the stop still counts)
3 + 0.5 + 1 = 4.5 h
3
average speed = total distance / total time
420 ÷ 4.5
✓93.3 km/h
A4
Convert 20 m/s into km/h.
1
Multiply by 3.6 (since 1 m/s = 3.6 km/h)
20 × 3.6
✓72 km/h
A5
Average speed 15 km/h over 45 minutes. Find distance.
1
Convert time to hours
45 min = 0.75 h
2
distance = speed × time
15 × 0.75
✓11.25 km
B. Distance-Time Graphs
B1
Line from (0 s, 0 m) to (8 s, 64 m). Find speed.
✓8 m/s
B2
At rest at 20 m (0-5 s), then 60 m by 15 s. Find speed while moving.
1
speed = change in distance / change in time (moving phase only)
(60 − 20) ÷ (15 − 5) = 40 ÷ 10
✓4 m/s
B3
Shape of distance-time graph for an object decelerating to rest.
✓A curve that starts steep and becomes progressively less steep, flattening into a horizontal line once the object stops (gradient, i.e. speed, continually decreasing to zero).
B4
Curve becoming steeper over time — what motion is this?
✓The object is accelerating (its speed, given by the gradient, is increasing).
C. Speed-Time Graphs and Area
C1
Constant 12 m/s for 6 s. Find distance.
1
distance = area under graph (rectangle)
12 × 6
✓72 m
C2
0 to 18 m/s in 4 s, then constant 18 m/s for 5 s. Find total distance.
1
Triangle area (0-4 s)
½ × 4 × 18 = 36 m
2
Rectangle area (4-9 s)
18 × 5 = 90 m
3
Add the two areas
36 + 90
✓126 m
C3
Decelerates 20 m/s to rest in 5 s. Find distance.
1
distance = area of triangle
½ × 5 × 20
✓50 m
C4
Finding distance from a curved (non-uniform acceleration) speed-time graph.
✓Since the area can’t be split into simple triangles/rectangles, count the number of grid squares under the curve (including part-squares, estimated) and multiply by the value each square represents, or use another appropriate approximation method — the area still equals the distance travelled.
D. Acceleration
D1
4 m/s to 22 m/s in 6 s. Find acceleration.
✓3 m/s²
D2
30 m/s to 10 m/s in 8 s. Find acceleration.
1
a = Δv/Δt
(10 − 30) ÷ 8 = −20 ÷ 8
✓−2.5 m/s² — a deceleration of 2.5 m/s²
D3
Line from (2s, 5 m/s) to (10s, 29 m/s). Find acceleration.
1
a = gradient
(29 − 5) ÷ (10 − 2) = 24 ÷ 8
✓3 m/s²
D4
Accelerates at 4 m/s² for 5 s from rest. Find final speed.
1
Rearrange a = Δv/Δt: final speed = initial speed + a×t
v = 0 + (4 × 5)
✓20 m/s
E. Free Fall and Terminal Velocity
E1
Falls freely for 3 s. Find speed gained. (g = 9.8 m/s²)
1
speed gained = g × t
9.8 × 3
✓29.4 m/s
E2
Falls freely from rest for 2.5 s. Find (a) speed, (b) distance fallen.
1
(a) speed = g × t
9.8 × 2.5 = 24.5 m/s
2
(b) distance = area under speed-time graph (triangle, from 0 to 24.5 m/s)
½ × 2.5 × 24.5
✓(a) 24.5 m/s (b) 30.625 m
E3
Why does a skydiver’s speed-time graph flatten near terminal velocity?
✓As speed increases, air resistance increases too, reducing the resultant force (and hence the acceleration, the graph’s gradient). Once air resistance equals weight, the resultant force and acceleration become zero, so the graph becomes a horizontal (flat) line at the constant terminal velocity.
Mastery Test — Answer Key
- 90 km/h (25 × 3.6)
- 60 km/h (90 ÷ 1.5)
- 9 m/s (45 ÷ 5)
- 24 m/s (v = u + at = 0 + 3×8)
- 120 m (10 × 12)
- 200 m (area of triangle: ½ × 10 × 40)
- 39.2 m/s (9.8 × 4)
- 4 m/s² — (23−3)/(6−1)
- Speed multiplied by time gives distance for constant speed; on a speed-time graph, this is exactly the area of the rectangle under the line. For changing speed, the same logic applies to each small time interval, so the total area under the whole graph — whatever its shape — equals the total distance travelled.
- 30 m — first triangle: ½×3×12=18 m; second triangle: ½×2×12=12 m; total = 18+12 = 30 m
