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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(edited 10 years ago)

1391342022033.jpg153.5KB

Reply 3

ghostwalker

17

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

\displaystyle a=\frac{dv}{dt}= \frac{dv}{dx}\times\frac{dx}{dt}= v\frac{dv}{dx}

Reply 4

cooldudeman

OP

18

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

\displaystyle a=\frac{dv}{dt}= \frac{dv}{dx}\times\frac{dx}{dt}= v\frac{dv}{dx}

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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Reply 5

ghostwalker

17

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.

Reply 6

brianeverit

9

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
Posted from TSR Mobile

acceleration = dv/dt or v.dv/dx. I didn't say anything about v.dv/dt

Reply 7

cooldudeman

OP

18

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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1391376209636.jpg68.4KB

Reply 8

brianeverit

9

Original post by cooldudeman

I did what you said but still didnt get an actual value.

Posted from TSR Mobile

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 -

Reply 9

cooldudeman

OP

18

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

Mech resistance q

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