The Student Room Group

Scroll to see replies

Original post by nukethemaly
Has anyone got a prediction paper they can post for C1?


What's a prediction paper?
Original post by joostan
Because the uu would be different.


ah yea that makes since

thanks man
Original post by TH3-FL45H
What's a prediction paper?


take a wild guess
Reply 63
Original post by upthegunners
ah yea that makes since

thanks man


no prob :smile:
Original post by TH3-FL45H
What's a prediction paper?


Someone called TheGoldenRatio uploaded one in Jan the night before the exam, so I was just wondering if someone's done anything like that again for this exam.
Reply 65
For C1 it's basically the same questions ever-so-slightly changed? Sometimes they're even in the same order. :tongue:
Original post by joostan
no prob :smile:


do you have any m1 tips? :smile:

I often find i over complicate things

I see an 8 mark question and just think it cannot be as simple as i think..

how do I over come this? thanks, oliver
Reply 67
Original post by upthegunners
do you have any m1 tips? :smile:

I often find i over complicate things

I see an 8 mark question and just think it cannot be as simple as i think..

how do I over come this? thanks, oliver


Diagram. Always draw a diagram with all the forces displayed, and work out what you're interested in.
DJMayes posted some tips, there's a link in the OP, I recommend you take a look :smile:
nerves have kicked in for M1 :frown:

Need help with this:

A particle P is moving so that its velocity, vms1v ms^{-1}, after tt seconds is given by v=3t34tv=3t^3 - 4t

Initially P is at rest and is a displacment of 3m from a fixed point O.

Find vv when t=1t=1.. okay so v=1v=-1.

ii) Find an expression for the displacment of P at any time tt.
d=(3t24t)dtd= \int (3t^2 - 4t)dt
d=t32t2+3d=t^3 - 2t^2 +3


Now this is where I need help!

Find the distance travelled by the particle before it returns to its intial position. thanks :smile:

EDIT: 69th post ROFL
(edited 10 years ago)
Reply 69
Original post by upthegunners
nerves have kicked in for M1 :frown:

Need help with this:

A particle P is moving so that its velocity, vms1v ms^{-1}, after tt seconds is given by v=3t34tv=3t^3 - 4t

Initially P is at rest and is a displacment of 3m from a fixed point O.

Find vv when t=1t=1.. okay so v=1v=-1.

ii) Find an expression for the displacment of P at any time tt.
d=(3t24t)dtd= \int (3t^2 - 4t)dt
d=t32t2+3d=t^3 - 2t^2 +3


Now this is where I need help!

Find the distance travelled by the particle before it returns to its intial position. thanks :smile:

EDIT: 69th post ROFL


Initial position => displacement = 0. Find t at this point to start with.
Original post by CD315
Initial position => displacement = 0. Find t at this point to start with.


So you set the expression for displacment equal to zero and solve?
Reply 71
anyone got any resources for c3 and c4? apart from solomon/elmwood ?
Reply 73
Original post by upthegunners
So you set the expression for displacment equal to zero and solve?


You made a mistake in your integration by the way - you integrated 3t23t^2 instead of 3t33t^3

EDIT: In my last post - I mean set the displacement equal to 3 instead of 0, as this is where it began.
(edited 10 years ago)
Reply 74
Original post by kashagupta
How would you find the area of region R?


Posted from TSR Mobile


Calculate area under triangle...

Then remove area under curve (integrate)
Reply 75
Original post by kashagupta
How would you find the area of region R?


Posted from TSR Mobile


Area of a triangle with third side directly below N, then subtract the area beneath the curve between M and directly below N.
Original post by CD315
You made a mistake in your integration by the way - you integrated 3t23t^2 instead of 3t33t^3

EDIT: In my last post - I mean set the displacement equal to 3 instead of 0, as this is where it began.


so far I have got:

when v=0 t=4/3 amd t=0

when t=0 d=3

when t=4/3 d=49/27

what other times do we take? thanks man
Original post by L'Evil Fish
Calculate area under triangle...

Then remove area under curve (integrate)


Ahh got it, thanks! Don't know why I just wasn't seeing it!


Posted from TSR Mobile
Original post by joostan
Area of a triangle with third side directly below N, then subtract the area beneath the curve between M and directly below N.


Thanks :smile:


Posted from TSR Mobile
Reply 79
Original post by upthegunners
so far I have got:

when v=0 t=4/3 amd t=0

when t=0 d=3

when t=4/3 d=49/27

what other times do we take? thanks man


From your integration, you should have d=34t42t2+3d=\dfrac{3}{4}t^4-2t^2+3

Now, find out the time at which the displacement is the same as it's original - i.e;

34t42t2+3=3\dfrac{3}{4}t^4-2t^2+3=3

When I solve this, I get t=83t=\sqrt{\dfrac{8}{3}} (correct me if I've made a mistake).

Latest

Trending

Trending