# Mech resistance q

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#1
Need help.on part 2b. How can you get a 'value'? Barely anything cancels.

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6 years ago
#2
(Original post by cooldudeman)
Need help.on part 2b. How can you get a 'value'? Barely anything cancels.

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Write acceleration as v.dv/dx and solve differential equation.
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#3
(Original post by brianeverit)
Write acceleration as v.dv/dx and solve differential equation.
Why is it a=v.dv/dt and not a=dv/dt?
Im guessing v=s.ds/dt as well
I even checked on this website.

Im assuming that the picture only applies when you have everything in terms of t right?

No one told me this stuff
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6 years ago
#4
(Original post by cooldudeman)
Why is it a=v.dv/dt and not a=dv/dt?
Read brianeverit's post again, he didn't say "v.dv/dt".

His suggestion is a standard technique introduced at A-level.

Comes from

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#5
(Original post by ghostwalker)
Read brianeverit's post again, he didn't say "v.dv/dt".

His suggestion is a standard technique introduced at A-level.

Comes from

Ohhhh okok thats not confusing anymore then.

So does that mean that I didn't do the first part completely wrong

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6 years ago
#6
(Original post by cooldudeman)
So does that mean that I didn't do the first part completely wrong
First part for time at max height was fine.
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6 years ago
#7
(Original post by cooldudeman)
Why is it a=v.dv/dt and not a=dv/dt?
Im guessing v=s.ds/dt as well
I even checked on this website.

Im assuming that the picture only applies when you have everything in terms of t right?

No one told me this stuff
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acceleration = dv/dt or v.dv/dx. I didn't say anything about v.dv/dt
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#8
(Original post by brianeverit)
acceleration = dv/dt or v.dv/dx. I didn't say anything about v.dv/dt
(Original post by ghostwalker)
First part for time at max height was fine.
I did what you said but still didnt get an actual value.

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6 years ago
#9
(Original post by cooldudeman)
I did what you said but still didnt get an actual value.

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By "actual value" , do you mean you are trying to get a single numerical value? I don't see how that would be possible. The height is bound to be some function of U and k. btw the second term in your first integration should be + not -
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#10
(Original post by brianeverit)
By "actual value" , do you mean you are trying to get a single numerical value? I don't see how that would be possible. The height is bound to be some function of U and k. btw the second term in your first integration should be + not -
Oh so the answer can be in terms of k and U. Ive never come across a question that said 'value' when it didn't mean a numerical one.

Yeah ill fix it
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