CIE iGCSE Co-Ordinated Science P1.2 Motion Exam Style Questions Paper 1
▶️ Answer/Explanation
From the graph, the car accelerates from rest and reaches a constant speed of 10 m/s at 10 seconds. After 10 seconds, the speed remains constant (the line is horizontal).
To find the distance travelled before reaching constant speed, we need to calculate the area under the speed-time graph from 0 to 10 seconds. This area is a triangle.
Area of a triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \)
\( \text{Area} = \frac{1}{2} \times 10 \times 10 = 50 \text{ m} \)
Therefore, the car travels a distance of 50 m before it reaches a constant speed.
✅ Answer: (B)
▶️ Answer/Explanation
Average speed is calculated using \(\text{average speed} = \dfrac{\text{total distance}}{\text{total time}}\).
The total distance travelled is \(100 + 300 = 400\text{ m}\).
The total time taken is \(10 + 15 = 25\text{ s}\).
So, average speed \(= \dfrac{400}{25} = 16\text{ m/s}\).
✅ Answer: (D)
The total distance travelled is \(100 + 300 = 400\text{ m}\).
The total time taken is \(10 + 15 = 25\text{ s}\).
So, average speed \(= \dfrac{400}{25} = 16\text{ m/s}\).
✅ Answer: (D)
▶️ Answer/Explanation
On a speed–time graph, the distance travelled equals the area under the graph line.
For an object accelerating uniformly from rest to speed \(v\) over time \(t\), this area is a triangle.
The area of this triangle is \(\frac{1}{2}vt\).
✅ Answer: (C)
For an object accelerating uniformly from rest to speed \(v\) over time \(t\), this area is a triangle.
The area of this triangle is \(\frac{1}{2}vt\).
✅ Answer: (C)
▶️ Answer/Explanation
At constant speed, distance increases proportionally with time, giving a straight diagonal line on the distance–time graph; simultaneously, speed stays constant, giving a horizontal line on the speed–time graph.
The car starts at point X (not at zero distance), so the distance–time graph starts above the origin — option A shows both these features correctly.
✅ Answer: (A)
The car starts at point X (not at zero distance), so the distance–time graph starts above the origin — option A shows both these features correctly.
✅ Answer: (A)
▶️ Answer/Explanation
The distance travelled is equal to the area under the speed–time graph, which here is a trapezium with parallel sides of speed 10 m/s and 18 m/s over 5 s.
Using the trapezium area formula: \( \text{distance} = \dfrac{1}{2}(10+18)\times 5 = \dfrac{1}{2}\times 28 \times 5 = 70 \) m.
So the car travels 70 m in the time shown.
✅ Answer: (C)
Using the trapezium area formula: \( \text{distance} = \dfrac{1}{2}(10+18)\times 5 = \dfrac{1}{2}\times 28 \times 5 = 70 \) m.
So the car travels 70 m in the time shown.
✅ Answer: (C)