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Question Number 9101 by tawakalitu last updated on 18/Nov/16
Commented by mrW last updated on 18/Nov/16
b) see a) below 15s+4=mv(dv/ds) (15s+4)ds=mvdv ∫(15s+4)ds=∫mvdv ((15)/2)s^2 +4s+C=(1/2)mv^2 at time 0: s=0, v=0 C=0 ((15)/2)s^2 +4s−(1/2)mv^2 =0 15s^2 +8s−mv^2 =0 v=(√((15s^2 +8s)/m)) s=((−8+(√(64+60mv^2 )))/(30)) (i): when s=2m, v=(√((15×2^2 +8×2)/1))=(√(76))≈8.72m/s (ii): when v=8m/s, s=((−8+(√(64+60×1×8^2 )))/(30))≈1.816m
$$\left.{b}\right) \\ $$$$\left.{see}\:{a}\right)\:{below} \\ $$$$\mathrm{15}{s}+\mathrm{4}={mv}\frac{{dv}}{{ds}} \\ $$$$\left(\mathrm{15}{s}+\mathrm{4}\right){ds}={mvdv} \\ $$$$\int\left(\mathrm{15}{s}+\mathrm{4}\right){ds}=\int{mvdv} \\ $$$$\frac{\mathrm{15}}{\mathrm{2}}{s}^{\mathrm{2}} +\mathrm{4}{s}+{C}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$${at}\:{time}\:\mathrm{0}:\:{s}=\mathrm{0},\:{v}=\mathrm{0} \\ $$$${C}=\mathrm{0} \\ $$$$\frac{\mathrm{15}}{\mathrm{2}}{s}^{\mathrm{2}} +\mathrm{4}{s}−\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{15}{s}^{\mathrm{2}} +\mathrm{8}{s}−{mv}^{\mathrm{2}} =\mathrm{0} \\ $$$${v}=\sqrt{\frac{\mathrm{15}{s}^{\mathrm{2}} +\mathrm{8}{s}}{{m}}} \\ $$$${s}=\frac{−\mathrm{8}+\sqrt{\mathrm{64}+\mathrm{60}{mv}^{\mathrm{2}} }}{\mathrm{30}} \\ $$$$\left({i}\right):\:{when}\:{s}=\mathrm{2}{m}, \\ $$$${v}=\sqrt{\frac{\mathrm{15}×\mathrm{2}^{\mathrm{2}} +\mathrm{8}×\mathrm{2}}{\mathrm{1}}}=\sqrt{\mathrm{76}}\approx\mathrm{8}.\mathrm{72}{m}/{s} \\ $$$$\left({ii}\right):\:{when}\:{v}=\mathrm{8}{m}/{s}, \\ $$$${s}=\frac{−\mathrm{8}+\sqrt{\mathrm{64}+\mathrm{60}×\mathrm{1}×\mathrm{8}^{\mathrm{2}} }}{\mathrm{30}}\approx\mathrm{1}.\mathrm{816}{m} \\ $$
Commented by tawakalitu last updated on 18/Nov/16
Wow, this is great. God bless you sir.
$$\mathrm{Wow},\:\mathrm{this}\:\mathrm{is}\:\mathrm{great}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mrW last updated on 19/Nov/16
a) at any time t: distance from point 0 is s(t), speed is v(t), acceleration is A(t). v=(ds/dt) A=(dv/dt)=(dv/ds)∙(ds/dt)=v∙(dv/ds) according to newton′s second law of motion: F=mA (1/(s+a))=mv(dv/ds) (1/(s+a))ds=mvdv ∫(1/(s+a))ds=∫mvdv ∫(1/(s+a))d(s+a)=m∫vdv ln (s+a)=(1/2)mv^2 +C s+a=e^(((1/2)mv^2 +C)) s=e^(((1/2)mv^2 +C)) −a=e^C ∙e^((1/2)mv^2 ) −a at time t=0: s=0, v=0 0=e^C −a e^C =a s=a(e^((1/2)mv^2 ) −1) continue...
$$\left.{a}\right) \\ $$$${at}\:{any}\:{time}\:{t}: \\ $$$${distance}\:{from}\:{point}\:\mathrm{0}\:{is}\:{s}\left({t}\right),\: \\ $$$${speed}\:{is}\:{v}\left({t}\right),\:{acceleration}\:{is}\:{A}\left({t}\right). \\ $$$${v}=\frac{{ds}}{{dt}} \\ $$$${A}=\frac{{dv}}{{dt}}=\frac{{dv}}{{ds}}\centerdot\frac{{ds}}{{dt}}={v}\centerdot\frac{{dv}}{{ds}} \\ $$$${according}\:{to}\:{newton}'{s}\:{second}\:{law}\:{of}\:{motion}: \\ $$$${F}={mA} \\ $$$$\frac{\mathrm{1}}{{s}+{a}}={mv}\frac{{dv}}{{ds}} \\ $$$$\frac{\mathrm{1}}{{s}+{a}}{ds}={mvdv} \\ $$$$\int\frac{\mathrm{1}}{{s}+{a}}{ds}=\int{mvdv} \\ $$$$\int\frac{\mathrm{1}}{{s}+{a}}{d}\left({s}+{a}\right)={m}\int{vdv} \\ $$$$\mathrm{ln}\:\left({s}+{a}\right)=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{C} \\ $$$${s}+{a}={e}^{\left(\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{C}\right)} \\ $$$${s}={e}^{\left(\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{C}\right)} −{a}={e}^{{C}} \centerdot{e}^{\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} } −{a} \\ $$$${at}\:{time}\:{t}=\mathrm{0}:\:{s}=\mathrm{0},\:{v}=\mathrm{0} \\ $$$$\mathrm{0}={e}^{{C}} −{a} \\ $$$${e}^{{C}} ={a} \\ $$$${s}={a}\left({e}^{\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} } −\mathrm{1}\right) \\ $$$$ \\ $$$${continue}… \\ $$
Commented by tawakalitu last updated on 18/Nov/16
This is nice. i really appreciate . God bless you sir.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{nice}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$$$\mathrm{sir}. \\ $$

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