Question Number 13563 by Tinkutara last updated on 21/May/17
Commented by Tinkutara last updated on 21/May/17
$$\mathrm{36}.\:\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{distance}\:{x}\:\mathrm{from}\:\mathrm{origin}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:{u}\:+\:{kx} \\ $$$$\left(\mathrm{2}\right)\:{u}\:−\:{kx} \\ $$$$\left(\mathrm{3}\right)\:{kx} \\ $$$$\left(\mathrm{4}\right)\:−{kx} \\ $$$$\mathrm{37}.\:\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:\mathrm{time} \\ $$$${t}\:\mathrm{after}\:\mathrm{start}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:{u}\:−\:{kt} \\ $$$$\left(\mathrm{2}\right)\:{ue}^{−{kt}} \\ $$$$\left(\mathrm{3}\right)\:{e}^{−{kt}} \\ $$$$\left(\mathrm{4}\right)\:{u}\:−\:{e}^{−{kt}} \\ $$
Commented by Tinkutara last updated on 21/May/17
Commented by Tinkutara last updated on 21/May/17
$$\mathrm{Answer}\:\mathrm{these}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{paragraph}. \\ $$
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) ] .](Q13567.png)
$$\left(\mathrm{36}\right). \\ $$$$\mathrm{a}=−\mathrm{kv} \\ $$$$\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}=−\mathrm{kv} \\ $$$$\int_{\boldsymbol{\mathrm{u}}} ^{\:\:\boldsymbol{\mathrm{v}}} \mathrm{dv}=−\mathrm{k}\int_{\mathrm{0}} ^{\:\:\boldsymbol{\mathrm{x}}} \mathrm{dx} \\ $$$$\mathrm{v}−\mathrm{u}=−\mathrm{kx} \\ $$$$\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{u}}−\boldsymbol{\mathrm{kx}}\:\:\:\:\:\:\:\left[\mathrm{option}\:\:\left(\mathrm{2}\right)\right]\:\:\:.....\left(\mathrm{i}\right) \\ $$$$ \\ $$$$\left(\mathrm{37}\right). \\ $$$$\mathrm{a}=−\mathrm{kv} \\ $$$$\frac{\mathrm{dv}}{\mathrm{dt}}=−\mathrm{kv} \\ $$$$\underset{\:\:\boldsymbol{\mathrm{u}}} {\int}^{\:\boldsymbol{\mathrm{v}}} \:\frac{\mathrm{dv}}{\mathrm{v}}=−\mathrm{k}\int_{\mathrm{0}} ^{\:\:\boldsymbol{\mathrm{t}}} \mathrm{dt} \\ $$$$\mathrm{ln}\:\left(\frac{\mathrm{v}}{\mathrm{u}}\right)=−\mathrm{kt} \\ $$$$\mathrm{v}=\mathrm{ue}^{−\boldsymbol{\mathrm{kt}}} \:\:\:\:\:\left[\mathrm{option}\:\:\left(\mathrm{2}\right)\:\right] \\ $$$$ \\ $$$$\left(\mathrm{38}\right). \\ $$$$\mathrm{a}=−\mathrm{kv} \\ $$$$\:\:=−\mathrm{k}\left(\mathrm{u}−\mathrm{kx}\right)\:\:\:\:\:\:\:\:\:\:..\mathrm{from}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{if}\:\mathrm{k}>\mathrm{0}\:,\:\mathrm{u}>\mathrm{0} \\ $$$$\mathrm{at}\:\mathrm{t}=\mathrm{0},\:\:\mathrm{a}=−\mathrm{ku} \\ $$$$\mathrm{a}=\mathrm{0}\:\:\mathrm{at}\:\:\mathrm{x}=\frac{\mathrm{u}}{\mathrm{k}} \\ $$$$\mathrm{so}\:\mathrm{a}-\mathrm{x}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{linear}\:\mathrm{with}\:\mathrm{negative} \\ $$$$\mathrm{a}\:\mathrm{intercept}\:\mathrm{and}\:\mathrm{positive} \\ $$$$\:\mathrm{x}-\mathrm{intercept}.\:\:\left[\mathrm{option}\:\left(\mathrm{3}\right)\:\right]\:. \\ $$
Commented by Tinkutara last updated on 21/May/17
$$\mathrm{Q}\:\mathrm{38}.\:\mathrm{Why}\:{k}\:>\:\mathrm{0}\:\mathrm{implies}\:{u}\:>\:\mathrm{0}?\:\mathrm{Answer} \\ $$$$\mathrm{38}\:\mathrm{is}\:\mathrm{not}\:\mathrm{understood}\:\mathrm{by}\:\mathrm{me}.\:\mathrm{Please} \\ $$$$\mathrm{explain}\:\mathrm{in}\:\mathrm{detail}\:\mathrm{Sir}. \\ $$
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) .](Q13571.png)
$$\mathrm{the}\:\mathrm{case}\:\mathrm{is}\:\mathrm{of}\:\mathrm{initial}\:\mathrm{speed}\:\mathrm{u}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{certain}\:\mathrm{direction},\:\mathrm{that}\:\mathrm{i}\:\mathrm{may}\:\mathrm{take} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{direction};\:\mathrm{so}\:\boldsymbol{\mathrm{u}}>\mathrm{0} \\ $$$$\mathrm{what}\:\mathrm{i}\:\mathrm{meant}\:\mathrm{that}\:\mathrm{if}\:\mathrm{u}>\mathrm{0}\:\mathrm{and} \\ $$$$\boldsymbol{\mathrm{k}}>\mathrm{0}\:\mathrm{both}\:\mathrm{then}\:\mathrm{object}'\mathrm{s}\:\mathrm{velocity}\:\mathrm{goes} \\ $$$$\mathrm{decreasing},\:\mathrm{but}\:\mathrm{is}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{and} \\ $$$$\mathrm{reaches}\:\mathrm{zero}\:\mathrm{in}\:\mathrm{infinite}\:\mathrm{time}\:; \\ $$$$\mathrm{the}\:\mathrm{position}\:\mathrm{coordinate}\:\mathrm{goes}\:\mathrm{near} \\ $$$$\mathrm{and}\:\mathrm{near}\:\mathrm{its}\:\mathrm{upper}\:\mathrm{limit}\:\mathrm{of}\:\mathrm{x}=\:\frac{\mathrm{u}}{\mathrm{k}}, \\ $$$$\mathrm{because}\:\mathrm{integrating}\:\mathrm{velocity} \\ $$$$\mathrm{x}=\int_{\mathrm{0}} ^{\:\:\boldsymbol{\mathrm{t}}} \mathrm{vdt}=\mathrm{u}\int_{\mathrm{0}} ^{\:\boldsymbol{\mathrm{t}}} \mathrm{e}^{−\boldsymbol{\mathrm{kt}}} \mathrm{dt} \\ $$$$\:\:\:=\mathrm{u}\left[\frac{\mathrm{e}^{−\boldsymbol{\mathrm{kt}}} }{−\mathrm{k}}\right]_{\boldsymbol{\mathrm{t}}=\mathrm{0}} ^{\boldsymbol{\mathrm{t}}=\boldsymbol{\mathrm{t}}} =\frac{\mathrm{u}}{\mathrm{k}}\left(\mathrm{1}−\mathrm{e}^{−\boldsymbol{\mathrm{kt}}} \right)\: \\ $$$$\mathrm{as}\:\mathrm{t}\rightarrow\infty\:\:,\:\mathrm{x}\rightarrow\frac{\mathrm{u}}{\mathrm{k}}\:. \\ $$
Commented by Tinkutara last updated on 22/May/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$