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# AQA MM2B - Variable Force differential equations Watch

1. Totally understood the theory and logic, but I think the questions in the book are purposely geared more towards the point that you need to select the correct variables for the problem:

A particle of mass m moves in a straight line under the action of a resistive force of magnitude ke^(v/u) where u and k are constants. When t=0 the speed of the particle is u.

Find the time during which the speed decreases to u/2.

Pretty sure it's just been convoluted with variables on purpose. Although all the questions are like this. I've breezed through the rest of the chapter but I'm suddenly utterly stumped....!

I know the answer is of the form t=f(u). I think the bit that's throwing me is the "when t=0, v=u" part. How/when/where I make use that fact?

Sorry long post. Cheers all!

Posted from TSR Mobile
2. (Original post by stuart_aitken)
Totally understood the theory and logic, but I think the questions in the book are purposely geared more towards the point that you need to select the correct variables for the problem:

A particle of mass m moves in a straight line under the action of a resistive force of magnitude ke^(v/u) where u and k are constants. When t=0 the speed of the particle is u.

Find the time during which the speed decreases to u/2.

Pretty sure it's just been convoluted with variables on purpose. Although all the questions are like this. I've breezed through the rest of the chapter but I'm suddenly utterly stumped....!

I know the answer is of the form t=f(u). I think the bit that's throwing me is the "when t=0, v=u" part. How/when/where I make use that fact?

Sorry long post. Cheers all!

Posted from TSR Mobile
u is the velocity when t = 0 when solving a differential equation a constant needs to be found. This fact will enable you to express the equation in terms of u.
3. (Original post by joostan)
u is the velocity when t = 0 when solving a differential equation a constant needs to be found. This fact will enable you to express the equation in terms of u.
Aaaaaaaah I knew it! OK I didn't know, haha, but there was something 'extra' that I knew I'd totally overlooked. All makes sense now.

Thanks bud

I always hated the '+c' in integration because it seemed so needless at that general level. Finally its use is apparent!

Posted from TSR Mobile
4. (Original post by stuart_aitken)
Aaaaaaaah I knew it! OK I didn't know, haha, but there was something 'extra' that I knew I'd totally overlooked. All makes sense now.

Thanks bud

I always hated the '+c' in integration because it seemed so needless at that general level. Finally its use is apparent!

Posted from TSR Mobile
No worries
5. (Original post by joostan)
No worries
OK I'm still lost.....

Any idea for this one? (Image attached)...

Best I can get is 4/(8t+1).....

Cheers...! Major rep when I'm on my computer

Sorry, it sounds like I'm just fishing for free answers now. I'm entirely self-taught so TSR is my only base for help...

Edit... DONE IT!

Cheers anyway though because I'm sure you'd have been able to help

Posted from TSR Mobile
6. (Original post by stuart_aitken)
OK I'm still lost.....

Any idea for this one? (Image attached)...

Best I can get is 4/(8t+1).....

Cheers...! Major rep when I'm on my computer

Sorry, it sounds like I'm just fishing for free answers now. I'm entirely self-taught so TSR is my only base for help...

Edit... DONE IT!

Cheers anyway though because I'm sure you'd have been able to help

Posted from TSR Mobile
Sorry, I wasn't paying attention. Well Done. Thumb's up

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