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Allisa
1 month ago
11

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

leration, and 5.0 s later the back of the first car passes you.
How long does it take after the train starts moving until the back of the seventh car passes you? All cars are the same length.

Express your answer with the appropriate units.

The solution HAS to include the Kinematic equations with constant acceleration

Please include a graph.

Physics
1 answer:
serg [3.5K]1 month ago
8 0
The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.
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1 month ago
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4 0
1 month ago
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serg [3582]

Answer:

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Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

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2 months ago
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