Qualification > Math
Additional Math Help HERE ONLY...!
J.Darren:
--- Quote from: cooldude on June 02, 2010, 04:53:31 pm ---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
--- End quote ---
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:
--- Quote from: J.Darren on June 02, 2010, 05:03:29 pm ---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 ...
--- End quote ---
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:
--- Quote from: jellybeans on June 02, 2010, 05:06:41 pm ---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?
--- End quote ---
I have no idea, but it seems that the marking scheme accepts both ...
jellybeans:
--- Quote from: J.Darren on June 02, 2010, 05:14:44 pm ---I have no idea, but it seems that the marking scheme accepts both ...
--- End quote ---
lol, okay then. thanks! :)
& thanks to cooldude aswellll, :) :)
... another one ... Q9iii) ? :-[
cooldude:
--- Quote from: J.Darren on June 02, 2010, 05:03:29 pm ---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 ...
--- End quote ---
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
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