4. (a) Which of these graphs represents an object moving with a constant velocity of 2 m/s? (1) (b) Figure 7 is a velocity/time graph showing a 34s part of a train's journey - Edexcel - GCSE Physics Combined Science - Question 4 - 2021 - Paper 1

To calculate the acceleration, use the formula:

$a = \frac{v - u}{t}$

Assuming the initial velocity ($u$) is 5 m/s and the final velocity ($v$) is 25 m/s after 34 s, we have:

$a = \frac{25 \text{ m/s} - 5 \text{ m/s}}{34 \text{ s}} = \frac{20 \text{ m/s}}{34 \text{ s}} \approx 0.588 \text{ m/s}^2$

Therefore, the acceleration of the train is approximately $0.59$ m/s².

The distance can be calculated using the area under the velocity-time graph, which is a trapezoid in this case. The formula to find the area of a trapezoid is:

$ext{Area} = \frac{1}{2} \times (b_1 + b_2) \times h$

Where:

• $b_1 = 5$ m/s (initial velocity),
• $b_2 = 25$ m/s (final velocity), and
• $h = 34$ s (time).

Thus, the area (distance) is:

$\text{Distance} = \frac{1}{2} \times (5 + 25) \times 34 = \frac{1}{2} \times 30 \times 34 = 510 \text{ m}$

The train travels a distance of 510 m.

During the first few seconds after take-off, the force produced by the rocket engines remains constant. Since acceleration is defined as the change in velocity over time ($a = \frac{F}{m}$), if the force is constant and the mass of the rocket remains nearly unchanged, the acceleration will also remain constant during this time. However, as fuel is consumed and the mass of the rocket decreases, the acceleration will increase due to the same constant force acting on a smaller mass, following Newton's second law.