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Question Number 13563 by Tinkutara last updated on 21/May/17

Commented by Tinkutara last updated on 21/May/17

36. The velocity of the particle at a distance x from origin is (1) u + kx (2) u − kx (3) kx (4) −kx 37. The velocity of the particle at time t after start is (1) u − kt (2) ue^(−kt) (3) e^(−kt) (4) u − e^(−kt)

36.Thevelocityoftheparticleatadistancexfromoriginis(1)u+kx(2)u−kx(3)kx(4)−kx37.Thevelocityoftheparticleattimetafterstartis(1)u−kt(2)ue−kt(3)e−kt(4)u−e−kt

Commented by Tinkutara last updated on 21/May/17

Commented by Tinkutara last updated on 21/May/17

Answer these questions from paragraph.

Answerthesequestionsfromparagraph.

Answered by ajfour last updated on 21/May/17

(36). a=−kv v(dv/dx)=−kv ∫_u ^( v) dv=−k∫_0 ^( x) dx v−u=−kx v=u−kx [option (2)] .....(i) (37). a=−kv (dv/dt)=−kv ∫^( v) _( u) (dv/v)=−k∫_0 ^( t) dt ln ((v/u))=−kt v=ue^(−kt) [option (2) ] (38). a=−kv =−k(u−kx) ..from (i) if k>0 , u>0 at t=0, a=−ku a=0 at x=(u/k) so a-x graph is linear with negative a intercept and positive x-intercept. [option (3) ] .

(36).a=−kvvdvdx=−kv∫uvdv=−k∫0xdxv−u=−kxv=u−kx[option(2)].....(i)(37).a=−kvdvdt=−kv∫uvdvv=−k∫0tdtln(vu)=−ktv=ue−kt[option(2)](38).a=−kv=−k(u−kx)..from(i)ifk>0,u>0att=0,a=−kua=0atx=uksoa−xgraphislinearwithnegativeainterceptandpositivex−intercept.[option(3)].

Commented by Tinkutara last updated on 21/May/17

Q 38. Why k > 0 implies u > 0? Answer 38 is not understood by me. Please explain in detail Sir.

Q38.Whyk>0impliesu>0?Answer38isnotunderstoodbyme.PleaseexplainindetailSir.

Commented by ajfour last updated on 21/May/17

the case is of initial speed u in a certain direction, that i may take to be the positive direction; so u>0 what i meant that if u>0 and k>0 both then object′s velocity goes decreasing, but is always positive and reaches zero in infinite time ; the position coordinate goes near and near its upper limit of x= (u/k), because integrating velocity x=∫_0 ^( t) vdt=u∫_0 ^( t) e^(−kt) dt =u[(e^(−kt) /(−k))]_(t=0) ^(t=t) =(u/k)(1−e^(−kt) ) as t→∞ , x→(u/k) .

thecaseisofinitialspeeduinacertaindirection,thatimaytaketobethepositivedirection;sou>0whatimeantthatifu>0andk>0boththenobject′svelocitygoesdecreasing,butisalwayspositiveandreacheszeroininfinitetime;thepositioncoordinategoesnearandnearitsupperlimitofx=uk,becauseintegratingvelocityx=∫0tvdt=u∫0te−ktdt=u[e−kt−k]t=0t=t=uk(1−e−kt)ast→∞,x→uk.

Commented by Tinkutara last updated on 22/May/17

Thanks Sir!

ThanksSir!

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