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Having a problem in steps while solving integrals

  • #1

Homework Statement


The problem is attached. I'm new to these problems (calculus).
I'm not getting my answer as any of the options. I need your help to know whether me or the slide is wrong.
?temp_hash=12720349bd7b2b5fb6e9c4da92df3e15.png




Homework Equations


x_x[/B]


The Attempt at a Solution


IMG_20180619_034801.jpg

Thank you for reading.
 

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Answers and Replies

  • #2
haruspex
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What is ∫v.dv?
 
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  • #3
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The slide is wrong in that x is not constant, k is constant. Your formulation is correct up to ∫ v dv as pointed out by haruspex.
 
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  • #4
What is ∫v.dv?
It is an equation to solve kinematic equations.
a=dv/dt 1
v=dx/dt 2
Divide 1 by 2
a/v= dv/dt⋅dt/dx
a/v=dv/dx
a⋅dx=v⋅dv
Integrate LHS and RHS and you get 3rd equation of motion:
0x∫a⋅dx=uv∫v⋅dv
a⋅x0x=v2|uv
ax-0=v2/2-u2/2
2ax=v2-u2
 
  • #5
The slide is wrong in that x is not constant, k is constant. Your formulation is correct up to ∫ v dv as pointed out by haruspex.
So, am I correct or not ? This was quite an ambiguous reply..
If not, can you point out the mistake clearly and correction. Please?
 
  • #6
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Your LHS (2/3) k^2 x^(3/2) is OK the RHS as you have corrected is v^2 / 2 with an upper limit of 2u and a lower limit of u.
 
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  • #7
haruspex
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=uv∫v⋅dv...=v2|uv
Quite so, but that does not seem to be what you wrote in your attachment in post #1. I do not see a 2 there.
 
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  • #8
Your LHS (2/3) k^2 x^(3/2) is OK the RHS as you have corrected is v^2 / 2 with an upper limit of 2u and a lower limit of u.
Quite so, but that does not seem to be what you wrote in your attachment in post #1. I do not see a 2 there.
So, I was incorrect in believing v to be a constant! Silly me. (I pondered over this single question for an hour and didn't get any sleep last night xD)
Thank you both for helping me. Have a great day/night.
 

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