anonymous
  • anonymous
The velocity of a particle ts after it starts from rest is vm/s,where v=1.25t-0.05t^2.Find i)the initial accelaration of the partcle ii)the displacement of the partcle from its staring point at the instant when its accelaration is 0.05m/s^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Callisto
  • Callisto
acceleration = (v-u) /t = [(1.25t - 0.05t^2) -0] / t = t(1.25 - 0.05t) /t = 1.25 - 0.05t ms^-2
anonymous
  • anonymous
ok
Callisto
  • Callisto
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 :(

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More answers

anonymous
  • anonymous
i think for the first the initial acceleration should be 1.25- 0.1t by doing differentitation of velocity u ll get acceleration
Callisto
  • Callisto
Trust heena, don't trust me (i'm serious)
anonymous
  • anonymous
nah it happens wid everyone @callissto but how can we find displacement now?
anonymous
  • anonymous
i did it like you calistro and i was wrong.Heena is right.
anonymous
  • anonymous
now at this stage t=0.because its at its initial accelaration.therefore a=1.25m/s^2
Callisto
  • Callisto
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.)
anonymous
  • anonymous
v^2-U^2=2as (1.25t-0.05t^2)^2=2x0.05S (1.25t-0.05t^2)^2/0.01=s
Callisto
  • Callisto
Once again , don't trust me :(
anonymous
  • anonymous
hey @callisto it happens wid all even wid me too several but the bests thing atleast we tried ok..
Callisto
  • Callisto
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) ?
anonymous
  • anonymous
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.. :?
anonymous
  • anonymous
one request from u plz kindly post the qn wid der respective group <3
Callisto
  • Callisto
nope, i was asking how do you know at the instant a = 0.05, v is equal to 1.25t-0.05t^2?
anonymous
  • anonymous
oh yess i m a big fool we have to do doubleintegration of acceleration to get displacement :P
anonymous
  • anonymous
its because this partcle is the same particle as for question 1 and also the only thing that has changed is accelaration
Callisto
  • Callisto
when acceleration changes, shouldn't the velocity also change?
anonymous
  • anonymous
S=0.025 t^2 do integration of acceleartion two time to get displacement thnx @callisto :)
Callisto
  • Callisto
One more question, can i do integration once of velocity to get displacement?
Callisto
  • Callisto
Note: it is a definite integral and you should find the respect time for it
anonymous
  • anonymous
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
anonymous
  • anonymous
calistro ya u can do that
Callisto
  • Callisto
right, i surrender :(
anonymous
  • anonymous
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
Callisto
  • Callisto
if v = constant, a = 0, that's what i know :(
anonymous
  • anonymous
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
anonymous
  • anonymous
agree wid CAllisto abu a=0
Callisto
  • Callisto
thanks for giving me a lesson :)
anonymous
  • anonymous
@callisto there is a file for u hope it helps
1 Attachment
Callisto
  • Callisto
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
anonymous
  • anonymous
yup u can :)
anonymous
  • anonymous
yep thanks guys
anonymous
  • anonymous
@a level student plz post this qn on physc sec nxt time plz .....
anonymous
  • anonymous
yeah ill do that

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