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acceleration = (v-u) /t = [(1.25t - 0.05t^2) -0] / t = t(1.25 - 0.05t) /t = 1.25 - 0.05t ms^-2
when a = 0.05 from (1) a=1.25 - 0.05t 0.05 = 1.25-0.05t -1.2 = -0.05t t = 24s By s= ut +(1/2)at^2 s = 0(24) + (1/2)(0.05)(24)^2 = do it yourself Honestly, i'm not sure if i have done (ii) correctly. I just guess it's like that... Sorry :(
i think for the first the initial acceleration should be 1.25- 0.1t by doing differentitation of velocity u ll get acceleration
Trust heena, don't trust me (i'm serious)
nah it happens wid everyone @callissto but how can we find displacement now?
i did it like you calistro and i was wrong.Heena is right.
now at this stage t=0.because its at its initial accelaration.therefore a=1.25m/s^2
By putting a= 0.05 to the equation to find out t, and then use integration ∫v dt from t=0 to t= (you found in acc.)
v^2-U^2=2as (1.25t-0.05t^2)^2=2x0.05S (1.25t-0.05t^2)^2/0.01=s
Once again , don't trust me :(
hey @callisto it happens wid all even wid me too several but the bests thing atleast we tried ok..
Sorry , just wanna ask how do you know the velocity 'at the instant when its accelaration is 0.05m/s^2' is =(1.25t-0.05t^2) ?
well i used the formula V^2-u^2=2as so i guess we have to do like this but abu t i m still not sure.. :?
one request from u plz kindly post the qn wid der respective group <3
nope, i was asking how do you know at the instant a = 0.05, v is equal to 1.25t-0.05t^2?
oh yess i m a big fool we have to do doubleintegration of acceleration to get displacement :P
its because this partcle is the same particle as for question 1 and also the only thing that has changed is accelaration
when acceleration changes, shouldn't the velocity also change?
S=0.025 t^2 do integration of acceleartion two time to get displacement thnx @callisto :)
One more question, can i do integration once of velocity to get displacement?
Note: it is a definite integral and you should find the respect time for it
No we are given constants on the v equation. yeah hennna.first weneed to find time on the first intergrated formula which leads to acclaration.(a=1.25-0.1t)a is 0.05 therefore t=12.Now we use limints for t as 12 and 0.to get 61.2m
calistro ya u can do that
right, i surrender :(
i dunno abu limits and all dat but i m ssure as in first we are said abu diferentiation and whenever we are said about instant we have to do integration i just cross chck my notes
if v = constant, a = 0, that's what i know :(
yes thats true and whenever u intergrate v=1.25t-0.05t^2 U intergrate with respect to t.For u to get a real number answer n its correct
agree wid CAllisto abu a=0
thanks for giving me a lesson :)
yup i know that , but still thanks :) i was think if i do integration of velocity instead of 2twice of accelertation, i can reduce the chance of making mistakes
yup u can :)
yep thanks guys
@a level student plz post this qn on physc sec nxt time plz .....
yeah ill do that