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Question-4624

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Question Number 4624 by sanusihammed last updated on 14/Feb/16
Answered by Yozzii last updated on 15/Feb/16
(12) The equation of motion of the particle is given as (d/ds)((v^2 /2))=−g−kv^2 . v is a function of s as s and v are related to each other by time t. ∴ (d/ds)((v^2 /2))=(1/2)×2v(dv/ds)=v(dv/ds). ∴ v(dv/ds)=−g−kv^2 =−(g+kv^2 ) (∗) At the maximum height s=h,v=0 and at v=u,s=0. ∴ Separating the variables of (∗) and integrating between the limits v=0,v=u and s=h,s=0, we get ∫_u ^0 (v/(g+kv^2 ))dv=∫_0 ^h −1ds (1/(2k))∫_u ^0 ((2kv)/(g+kv^2 ))dv=−s∣_0 ^h (k≠0) (1/(2k))log_e (g+kv^2 )∣_u ^0 =−h=0 h=((−1)/(2k))(log_e (g)−log_e (g+ku^2 )) h=(1/(2k))log_e (((g+ku^2 )/g)) (k≠0,g+ku^2 >0⇒k>−g/u^2 ) h=(1/(2k))log_e (1+((ku^2 )/g)) (13)(i) Let a be the acceleration of the particle in ms^(−2) . a is given as a function of time t in the form a=t+3. Since a=(dv/dt)⇒∫_v_0 ^v dv=∫_t_0 ^t adt ⇒v∣_v_0 ^v =∫_t_0 ^t (t+3)dt v−v_0 =0.5t^2 +3t∣_t_0 ^t v=v_0 +0.5t^2 +3t−0.5t_0 ^2 −3t_0 At t_0 =0,v_0 =2. ∴v=2+3t+0.5t^2 −0−0 v=2+3t+(1/2)t^2 (t≥0). At t=2, v=2+3×2+(1/2)×4=10ms^(−1) (ii) Since v=(ds/dt)⇒∫_s_0 ^s ds=∫_t_0 ^t vdt. At t_0 =0,s_0 =0. ∴∫_0 ^s ds=∫_0 ^t (2+3t+(1/2)t^2 )dt s∣_0 ^s =2t+(3/2)t^2 +(1/6)t^3 ∣_0 ^t s−0=2t+(3/2)t^2 +(1/6)t^3 −0 s=t(2+(3/2)t+(1/6)t^2 ) (t≥0) s is an increasing function for all t≥0 since (ds/dt)=v=2+3t+0.5t^2 ≥2+0+0=2>0. The particle moves along a straight line so that since s≥0, the distance=displacement. At t=2 s,s(2)=2(2+3+(2/3))=((34)/3)m. At t=4 s, s(4)=4(2+6+(8/3))=((128)/3) m ∴ required distance=s(4)−s(2)=((128−34)/3)=((94)/3)m
$$\left(\mathrm{12}\right)\:{The}\:{equation}\:{of}\:{motion}\:{of}\:{the}\:{particle} \\ $$$${is}\:{given}\:{as} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{ds}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=−{g}−{kv}^{\mathrm{2}} . \\ $$$${v}\:{is}\:{a}\:{function}\:{of}\:{s}\:{as}\:{s}\:{and}\:{v}\:{are}\: \\ $$$${related}\:{to}\:{each}\:{other}\:{by}\:{time}\:{t}. \\ $$$$\therefore\:\frac{{d}}{{ds}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{v}\frac{{dv}}{{ds}}={v}\frac{{dv}}{{ds}}. \\ $$$$\therefore\:{v}\frac{{dv}}{{ds}}=−{g}−{kv}^{\mathrm{2}} =−\left({g}+{kv}^{\mathrm{2}} \right)\:\:\:\:\:\:\left(\ast\right) \\ $$$${At}\:{the}\:{maximum}\:{height}\:{s}={h},{v}=\mathrm{0}\:{and} \\ $$$${at}\:{v}={u},{s}=\mathrm{0}. \\ $$$$\therefore\:{Separating}\:{the}\:{variables}\:{of}\:\left(\ast\right)\:{and} \\ $$$${integrating}\:{between}\:{the}\:{limits}\: \\ $$$${v}=\mathrm{0},{v}={u}\:{and}\:{s}={h},{s}=\mathrm{0},\:{we}\:{get} \\ $$$$\int_{{u}} ^{\mathrm{0}} \frac{{v}}{{g}+{kv}^{\mathrm{2}} }{dv}=\int_{\mathrm{0}} ^{{h}} −\mathrm{1}{ds} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}}\int_{{u}} ^{\mathrm{0}} \frac{\mathrm{2}{kv}}{{g}+{kv}^{\mathrm{2}} }{dv}=−{s}\mid_{\mathrm{0}} ^{{h}} \:\:\:\:\left({k}\neq\mathrm{0}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left({g}+{kv}^{\mathrm{2}} \right)\mid_{{u}} ^{\mathrm{0}} =−{h}=\mathrm{0} \\ $$$${h}=\frac{−\mathrm{1}}{\mathrm{2}{k}}\left({log}_{{e}} \left({g}\right)−{log}_{{e}} \left({g}+{ku}^{\mathrm{2}} \right)\right) \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left(\frac{{g}+{ku}^{\mathrm{2}} }{{g}}\right)\:\:\left({k}\neq\mathrm{0},{g}+{ku}^{\mathrm{2}} >\mathrm{0}\Rightarrow{k}>−{g}/{u}^{\mathrm{2}} \right) \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left(\mathrm{1}+\frac{{ku}^{\mathrm{2}} }{{g}}\right)\:\:\: \\ $$$$ \\ $$$$\left(\mathrm{13}\right)\left({i}\right)\:{Let}\:{a}\:{be}\:{the}\:{acceleration}\:{of}\:{the}\: \\ $$$${particle}\:{in}\:{ms}^{−\mathrm{2}} .\:{a}\:{is}\:{given}\:{as}\:{a}\:{function} \\ $$$${of}\:{time}\:{t}\:{in}\:{the}\:{form}\:{a}={t}+\mathrm{3}.\: \\ $$$${Since}\:{a}=\frac{{dv}}{{dt}}\Rightarrow\int_{{v}_{\mathrm{0}} } ^{{v}} {dv}=\int_{{t}_{\mathrm{0}} } ^{{t}} {adt} \\ $$$$\Rightarrow{v}\mid_{{v}_{\mathrm{0}} } ^{{v}} =\int_{{t}_{\mathrm{0}} } ^{{t}} \left({t}+\mathrm{3}\right){dt} \\ $$$${v}−{v}_{\mathrm{0}} =\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}{t}\mid_{{t}_{\mathrm{0}} } ^{{t}} \\ $$$${v}={v}_{\mathrm{0}} +\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{0}.\mathrm{5}{t}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{3}{t}_{\mathrm{0}} \\ $$$${At}\:{t}_{\mathrm{0}} =\mathrm{0},{v}_{\mathrm{0}} =\mathrm{2}. \\ $$$$\therefore{v}=\mathrm{2}+\mathrm{3}{t}+\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} −\mathrm{0}−\mathrm{0} \\ $$$${v}=\mathrm{2}+\mathrm{3}{t}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \:\:\:\:\left({t}\geqslant\mathrm{0}\right). \\ $$$${At}\:{t}=\mathrm{2},\:{v}=\mathrm{2}+\mathrm{3}×\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}=\mathrm{10}{ms}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{Since}\:{v}=\frac{{ds}}{{dt}}\Rightarrow\int_{{s}_{\mathrm{0}} } ^{{s}} {ds}=\int_{{t}_{\mathrm{0}} } ^{{t}} {vdt}. \\ $$$${At}\:{t}_{\mathrm{0}} =\mathrm{0},{s}_{\mathrm{0}} =\mathrm{0}. \\ $$$$\therefore\int_{\mathrm{0}} ^{{s}} {ds}=\int_{\mathrm{0}} ^{{t}} \left(\mathrm{2}+\mathrm{3}{t}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt} \\ $$$${s}\mid_{\mathrm{0}} ^{{s}} =\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}} \mid_{\mathrm{0}} ^{{t}} \\ $$$${s}−\mathrm{0}=\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}} −\mathrm{0} \\ $$$${s}={t}\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}{t}+\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{2}} \right)\:\:\left({t}\geqslant\mathrm{0}\right) \\ $$$${s}\:{is}\:{an}\:{increasing}\:{function}\:{for}\:{all}\:{t}\geqslant\mathrm{0} \\ $$$${since}\:\frac{{ds}}{{dt}}={v}=\mathrm{2}+\mathrm{3}{t}+\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} \geqslant\mathrm{2}+\mathrm{0}+\mathrm{0}=\mathrm{2}>\mathrm{0}. \\ $$$${The}\:{particle}\:{moves}\:{along}\:{a}\:{straight}\: \\ $$$${line}\:{so}\:{that}\:{since}\:{s}\geqslant\mathrm{0},\:{the}\:{distance}={displacement}. \\ $$$${At}\:{t}=\mathrm{2}\:{s},{s}\left(\mathrm{2}\right)=\mathrm{2}\left(\mathrm{2}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{34}}{\mathrm{3}}{m}. \\ $$$${At}\:{t}=\mathrm{4}\:{s},\:{s}\left(\mathrm{4}\right)=\mathrm{4}\left(\mathrm{2}+\mathrm{6}+\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\mathrm{128}}{\mathrm{3}}\:{m}\: \\ $$$$\therefore\:{required}\:{distance}={s}\left(\mathrm{4}\right)−{s}\left(\mathrm{2}\right)=\frac{\mathrm{128}−\mathrm{34}}{\mathrm{3}}=\frac{\mathrm{94}}{\mathrm{3}}{m} \\ $$$$ \\ $$

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