6 (a) Which of these graphs represents an object moving with a constant velocity of 2 m/s? (b) Figure 8 is a velocity/time graph showing a 34s part of a train's journey - Edexcel - GCSE Physics - Question 6 - 2021 - Paper 1

To determine the acceleration, we use the formula: ext{acceleration} = rac{ ext{change in velocity}}{ ext{time}} From the graph, the change in velocity can be seen as moving from 0 m/s to 30 m/s over 34 seconds. Thus: ext{acceleration} = rac{30 ext{ m/s} - 0 ext{ m/s}}{34 ext{ s}} = rac{30}{34} ext{ m/s}^2 \ ext{acceleration} \approx 0.88 ext{ m/s}^2 Therefore, the acceleration of the train in the 34s is approximately 0.88 m/s².

The distance can be calculated using the area under the velocity-time graph. The graph shows a triangular area, which can be calculated using: ext{Area} = rac{1}{2} imes ext{base} imes ext{height} Here, the base is 34 seconds and the height is 30 m/s. Thus: ext{Distance} = rac{1}{2} imes 34 imes 30 = 510 ext{ m} Hence, the distance the train travels in 34s is 510 m.

During the first few seconds after take-off, the acceleration of the rocket increases. This is because the net force acting on the rocket is determined by Newton's second law, $F = ma$. As the rocket engines produce a constant thrust force, the mass of the rocket decreases due to burning fuel, therefore increasing acceleration. The relationship can be defined as: a = rac{F}{m} As the mass decreases while the force remains constant, the acceleration becomes greater.