4. (a) Which of these graphs represents an object moving with a constant velocity of 2 m/s? (a) A (b) B (c) C (d) D (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 can use the formula:

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

From the graph, the initial velocity (u) is 5 m/s and the final velocity (v) is 20 m/s over a time interval (t) of 34 s. Therefore,

$a = \frac{20 m/s - 5 m/s}{34 s} = \frac{15 m/s}{34 s} \approx 0.44 m/s^2$

So, the acceleration of the train in the 34s is approximately 0.44 m/s².

The distance can be found using the area under the velocity-time graph. The graph shows a trapezoid; thus:

Using the formula for the area of a trapezoid:

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

Where:

• $b_1$ = initial velocity = 5 m/s
• $b_2$ = final velocity = 20 m/s
• $h$ = time = 34 s

Simplifying gives:

$\text{Area} = \frac{1}{2} \times (5 m/s + 20 m/s) \times 34 s = \frac{1}{2} \times 25 m/s \times 34 s = 425 m$

Thus, the distance the train travels in the 34 seconds is 425 m.

During the first few seconds after take-off, the force produced by the rocket engines remains constant. Since acceleration is calculated using Newton's second law, where

$F = m \times a$

If force (F) remains constant and the mass (m) remains unchanged at the initial phase of the rocket's flight, the acceleration (a) will remain constant as well. Therefore, the rocket experiences a uniform acceleration during these initial seconds.