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 find the acceleration, we first need to determine the change in velocity over the given time. By analyzing Figure 7, we note the initial and final velocities. The acceleration can be calculated using the formula:

$a = \frac{v_f - v_i}{t}$

Assuming initial velocity $v_i = 5 m/s$ and final velocity $v_f = 20 m/s$, and following this formula:

$a = \frac{20 \text{ m/s} - 5 \text{ m/s}}{34 \text{ s}} = \frac{15}{34} \approx 0.44 \text{ m/s²}$

Thus, the acceleration is approximately 0.44 m/s² (to 2 significant figures).

To calculate the distance, we can use the area under the velocity-time graph. The area can be found using the equation for the area of a triangle plus the area of a rectangle:

• The triangle area (from $t=0$ to $t=20s$) is: $Area_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 20 \times 20 = 200 \text{ m}$
• The rectangle area (from $t=20s$ to $t=34s$, velocity = 20 m/s) is: $Area_{rectangle} = base \times height = 14 \times 20 = 280 \text{ m}$
• Total distance is: $Distance = 200 + 280 = 480 \text{ m}$.

During the first few seconds after take-off, the rocket experiences a constant force due to the rocket engines. According to Newton's second law, the acceleration of the rocket is directly proportional to the force applied and inversely proportional to its mass:

$F = ma \Rightarrow a = \frac{F}{m}$

Since the force remains constant and the mass of the rocket is decreasing (due to fuel consumption), the acceleration will increase during the first few seconds after take-off.