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A-particle-P-moves-in-a-plane-such-that-at-time-t-seconds-its-velocity-v-2ti-t-3-ms-1-a-Find-when-t-2-the-magnitudeof-the-i-velocity-of-P-ii-acceleration-of-P-b-Given-that-P-is-a




Question Number 173293 by pete last updated on 09/Jul/22
A particle P moves in a plane such that  at time t seconds, its velocity, v=(2ti−t^3 )ms^(−1) .  (a) Find, when t=2, the magnitudeof the:  (i) velocity of P.  (ii) acceleration of P.  (b) Given that P is at the point with position   vector (3i+2j) when t=1, find the position  vector of P when t=2.
AparticlePmovesinaplanesuchthatattimetseconds,itsvelocity,v=(2tit3)ms1.(a)Find,whent=2,themagnitudeofthe:(i)velocityofP.(ii)accelerationofP.(b)GiventhatPisatthepointwithpositionvector(3i+2j)whent=1,findthepositionvectorofPwhent=2.
Answered by Ar Brandon last updated on 09/Jul/22
v=(2ti−t^3 j)ms^(−1)   a.  i- v_(t=2) =∣4i−8j∣=(√(4^2 +8^2 ))=4(√5)ms^(−1)   ii- a(t)=v′(t)=(d/dt)(2ti−t^3 j)=(2i−3t^2 j)ms^(−2)          a_(t=2) =∣2i−12j∣=(√(148))ms^(−2) =2(√(37))ms^(−2)     b.  P(t)=∫v(t)dt=∫(2ti−t^3 j)dt=t^2 i−(t^4 /4)j+C  P(1)=3i+2j= i−(1/4)j+C⇒C=2i+(9/4)j  P(t)=(t^2 +2)i−(1/4)(t^4 −9)j  P(2)=6i−(7/4)j
v=(2tit3j)ms1a.ivt=2=∣4i8j∣=42+82=45ms1iia(t)=v(t)=ddt(2tit3j)=(2i3t2j)ms2at=2=∣2i12j∣=148ms2=237ms2b.P(t)=v(t)dt=(2tit3j)dt=t2it44j+CP(1)=3i+2j=i14j+CC=2i+94jP(t)=(t2+2)i14(t49)jP(2)=6i74j
Commented by pete last updated on 09/Jul/22
Thank you sir, I am grateful.
Thankyousir,Iamgrateful.
Commented by Tawa11 last updated on 11/Jul/22
Great sir
Greatsir

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