Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$$\mathrm{This}\:\mathrm{is}\:\mathrm{nice}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$$$\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com