Additional Math Help HERE ONLY...!

[cooldude]

HIIII, um, i need help with Q12ii) pleaseeee
Thanks

i knw you integrate the velocity to get the displacement but what values do you sub in? 
thanks again. XD

ii) integrate a to find v->
v=1.4t-0.3t^2+k
v=0.5, when t=0, thus k=0.5
thus v=1.4t-0.3t^2+0.5
x=0.7t^2-0.1t^3+0.5t
to find the dist u just have to integrate between the limits, 10 and 0, u dont have to worry about the arbitary constant as when we evaluate definite integrals we dont consider it, so now just substitute the value of 10 in the equation for x as substituting 0 wud give the displacement as 0,
x=70-100+5=-25m

[jellybeans]

ii) integrate a to find v->
v=1.4t-0.3t^2+k
v=0.5, when t=0, thus k=0.5
thus v=1.4t-0.3t^2+0.5
x=0.7t^2-0.1t^3+0.5t
to find the dist u just have to integrate between the limits, 10 and 0, u dont have to worry about the arbitary constant as when we evaluate definite integrals we dont consider it, so now just substitute the value of 10 in the equation for x as substituting 0 wud give the displacement as 0,
x=70-100+5=-25m

mmm thats what i did in the first place & got the same answer. but the MS had a different answer ... explanation ? hehs
*whats with the 2x7.5 ?  

thanks

[J.Darren]

HIIII, um, i need help with Q12ii) pleaseeee
Thanks

i knw you integrate the velocity to get the displacement but what values do you sub in?  
thanks again. XD

When you are dealing with displacement question, one must be very careful with the velocity, if between the given time frame, the particle experiences a change in the direction of travel (forwards / backwards), then you must do partial integration.

[jellybeans]

When you are dealing with displacement question, one must be very careful with the velocity, if between the given time frame, the particle experiences a change in the direction of travel (forwards / backwards), then you must do partial integration.

ohohhh, but why is it 2x7.5? and not just 7.5+25?

[cooldude]

mmm thats what i did in the first place & got the same answer. but the MS had a different answer ... explanation ? hehs
*whats with the 2x7.5 ? 

thanks

k sorry, its about the particle travelling 7.5m in the positive direction and then going back thus we multiply by 2 and then add the 25 m in the negative direction to find the total dist travelled, first the particle travels in the positive x direction and travels 7.5 m, and then for the next 5 seconds it travels 7.5 m back to the origin and then another 25 m in the negative x direction, thus a total distance of 7.5+7.5+25=40 m, sorry i misread the q, i thought it was askin the displacement not the distance travelled, neway j darren answered ur q

[J.Darren]

k sorry, its about the particle travelling 7.5m in the positive direction and then going back thus we multiply by 2 and then add the 25 m in the negative direction to find the total dist travelled, first the particle travels in the positive x direction and travels 7.5 m, and then for the next 5 seconds it travels 7.5 m back to the origin and then another 25 m in the negative x direction, thus a total distance of 7.5+7.5+25=40 m, sorry i misread the q, i thought it was askin the displacement not the distance travelled, neway j darren answered ur q

The question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...

[jellybeans]

The question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...

so why do you have to times 7.5 by twooooo? .. i mean how* do you know you have to times it by 2...
is it cuz it travels and then stops.. so you double the distance?

[J.Darren]

so why do you have to times 7.5 by twooooo? .. i mean how* do you know you have to times it by 2...
is it cuz it travels and then stops.. so you double the distance?

I have no idea, but it seems that the marking scheme accepts both ...

[jellybeans]

I have no idea, but it seems that the marking scheme accepts both ...

lol, okay then. thanks!
& thanks to cooldude aswellll,

... another one ... Q9iii) ? 

[cooldude]

The question did not mention anything the particles going back after travelling 7.5m in the positive direction, the acceleration gradually slows down after passing through O until it becomes zero at five, from that point onwards the particle is travelling in the negative direction until t = 10 ...

actually the q wont mention that the displacement is 7.5m u have to find this by calculation, ill show u what i mean, at t=5 sec the displacement is +7.5m
x=0.7t^2-0.1t^3+0.5t
when t=5, x=+7.5m
then the displacement is 0 at-->
0=0.7t^2-0.1t^3+0.5t
at t=7.65 the displacement is again 0 proving that the particle did travel +7.5m in the positive direction between 0 and 5, and then between 0 and 7.65 it again travels back to the origin a further 7.5m, the dist travelled=15m
now for the last 2.35 sec, u can find the displacement, u can integrate between 10 and 7.65 in the displacement, we get -25m, proving that the total dist is 7.5+7.5+25=40 , this should clear things up, if u need more explanation ill be happy to give it

[jellybeans]

actually the q wont mention that the displacement is 7.5m u have to find this by calculation, ill show u what i mean, at t=5 sec the displacement is +7.5m
x=0.7t^2-0.1t^3+0.5t
when t=5, x=+7.5m
then the displacement is 0 at-->
0=0.7t^2-0.1t^3+0.5t
at t=7.65 the displacement is again 0 proving that the particle did travel +7.5m in the positive direction between 0 and 5, and then between 0 and 7.65 it again travels back to the origin a further 7.5m, the dist travelled=15m
now for the last 2.35 sec, u can find the displacement, u can integrate between 10 and 7.65 in the displacement, we get -25m, proving that the total dist is 7.5+7.5+25=40 , this should clear things up, if u need more explanation ill be happy to give it

so you have to find out t when displacement = 0 ? ..which will then tell you that it returns to its origin.. and so .. its 7.5 x 2 for going there and back?
 

[J.Darren]

ln y = ln a + x ln b

Y = ln y
X = x
m = ln b
c = ln a

ln y = ln A + k ln x

Y = ln y
X = ln x
m = k
c = ln a

[jellybeans]

ln y = ln a + x ln b

Y = ln y
X = x
m = ln b
c = ln a

ln y = ln A + k ln x

Y = ln y
X = ln x
m = k
c = ln a

wait sorry.. is that the 3rd bit? :O
cuz for the first two i used logs instead of ln , ...  

anyways; for iii) i did:

px+qy=xy

xy-qy = px
(x-q)y = px
y = px / (x-q)

lol .. and yh. thats how far i got for 9iii) Xb

[cooldude]

so you have to find out t when displacement = 0 ? ..which will then tell you that it returns to its origin.. and so .. its 7.5 x 2?
 

yeah, if u get more than one value it basically tells u that the particle passes the origin at more than one point, i.e. it moves first in the positive direction and then in the negative direction, and then u have to find the distance travelled in the positive and the negative direction, and theres also one more way to find if the particle goes in the positive direction, if u find the maximum displacement and it comes to a value not equal to ur final displacement, i.e. in this q if the maximum displacement is not -25m then u get to knw that the particle has travelled in the positive direction also, for example in this u get the maximum displacement at t=5, i.e. x=7.5m, thus u get to knw that the particle has also travelled in the positive direction and u can take the movement in the positive direction into account

[J.Darren]

wait sorry.. is that the 3rd bit? :O
cuz for the first two i used logs instead of ln , ... 

px+qy=xy

xy-qy = px
(x-q)y = px
y = px / (x-q)

lol .. and yh. thats how far i got for 9iii) Xb

x - q = px / y

(x - q) / x = (px / y) * (1 / x)

(x - q) / x = p / y

1 - q / x = p / y

y = p / (1 - q / x)

would it works ?