Photo AI

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

Question icon

Question 6

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.png

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'... show full transcript

Worked Solution & Example Answer: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

Step 1

Which of these graphs represents an object moving with a constant velocity of 2 m/s?

96%

114 rated

Answer

Graph A is the correct representation of an object moving with a constant velocity of 2 m/s. In a distance-time graph, a straight line with a positive slope indicates uniform motion.

Step 2

i) Calculate the acceleration of the train in the 34 s.

99%

104 rated

Answer

To calculate the acceleration, we first need to determine the change in velocity and the time period.

From Figure 8, the initial velocity at 0 s is 0 m/s and the final velocity at 34 s is 25 m/s.

Using the formula for acceleration:

a=ΔvΔta = \frac{\Delta v}{\Delta t}
where ( \Delta v = v_{final} - v_{initial} = 25 - 0 = 25 ) m/s
and ( \Delta t = 34 ) s,

we find:

a=25m/s34s0.735 m/s2a = \frac{25 \, \text{m/s}}{34 \, \text{s}} \approx 0.735 \text{ m/s}^2

Rounding to 3 significant figures, the acceleration is approximately 0.735 m/s².

Step 3

ii) Calculate the distance the train travels in the 34 s.

96%

101 rated

Answer

To calculate the distance traveled, we can use the formula for distance in uniformly accelerated motion:

s=ut+12at2s = ut + \frac{1}{2}at^2
where ( u = 0 \text{ m/s} ) (initial velocity), ( a \approx 0.735 \text{ m/s}^2 ), and ( t = 34 \text{ s} ).

Plugging in the values, we get:

s=034+120.735(34)20+120.7351156425.82ms = 0 \cdot 34 + \frac{1}{2} \cdot 0.735 \cdot (34)^2 \approx 0 + \frac{1}{2} \cdot 0.735 \cdot 1156 \approx 425.82 \, \text{m}

Therefore, the distance the train travels in 34 s is approximately 426 m (to 3 significant figures).

Step 4

Explain what happens to the acceleration during the first few seconds.

98%

120 rated

Answer

During the first few seconds after take-off, the force produced by the rocket engines remains constant. According to Newton's second law, the relationship between force, mass, and acceleration is given by:

F=maF = ma
If the force (F) remains constant and the mass (m) of the rocket decreases as fuel is burnt, then the acceleration (a) will increase. This is because as mass decreases for a constant force, the acceleration must increase to satisfy the equation, leading to a greater rate of change in velocity.

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;