What is the speed of a speed skater at the finish line of a downhill course?

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SUMMARY

The speed skater in the problem accelerates uniformly from an initial speed of 30 m/s over a 200m downhill course, taking 5 seconds to finish. The calculations show that her final speed at the finish line is 50 m/s, derived from the equation s = 1/2 (u+v)t. The discrepancy with the book's answer of 45 m/s is highlighted, as the average speed of 37.5 m/s does not support the distance covered in the given time. The conclusion is that the published answer is incorrect.

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  • Understanding of kinematic equations, specifically s = 1/2 (u+v)t and Vf = V0 + a*t.
  • Knowledge of uniform acceleration concepts in physics.
  • Ability to manipulate algebraic equations to solve for unknown variables.
  • Familiarity with the concept of average speed and its calculation.
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  • Review kinematic equations for uniformly accelerated motion.
  • Study examples of uniform acceleration problems in physics textbooks.
  • Practice calculating final velocities using different initial speeds and distances.
  • Explore common misconceptions in physics problem-solving, particularly in kinematics.
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Students studying physics, particularly those focusing on kinematics and acceleration, as well as educators looking to clarify common errors in problem-solving approaches.

jaxreid
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Homework Statement


A speed skater crosses the start line of a straight 200m downhiull course with speed of 30m per second. She accellerated uniformly all the way down taking 5 sec to cover the course - what is her speed at the finish line? I know this should be simple, but just can't get the answer to work out the same as the solution in the paper.


Homework Equations



s = 1/2 (u+v)t

The Attempt at a Solution


200 = 1/2 (30+v)5
200/5 = 1/2(30+v)
40 = 1/2(30+v)
2 x 40 = 30 + v
80 - 30 = v
v = 50m per sec

Answer in book is 45 - HOW?
 
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I have to wonder about the answer you are given. According to them the average speed is .5(30+45)=37.5 m/s. That won't cover the 200 m in 5 seconds. I agree with you.

The constant (uniform) acceleration rate is 4 m/s*s gotten from: x = V0*t+.5*a*t^2

Inserting that into Vf = V0 + a*t yields Vf=50 m/s.

So I don't understand either.
 
Lawrence - many thanks - was starting to think I'd lost the plot, am glad you agree. The question's from a past paper and the published answer was 45. Just goes to show, they don't always get it right, will point this out to my physics teacher when school returns on Tuesday. Thanks again for your prompt reply.

Jax
 

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