Vector Questions

Question Number 10528 by Saham last updated on 16/Feb/17

$$\mathrm{Give}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{field} \\$$$$\mathrm{v}\:=\:\left(\mathrm{6}\:+\:\mathrm{2xy}\:+\:\mathrm{t}^{\mathrm{2}} \right)\mathrm{i}\:−\:\left(\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{10t}\right)\mathrm{j}\:+\:\mathrm{25k} \\$$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:\left(\mathrm{3},\:\mathrm{0},\:\mathrm{2}\right) \\$$$$\mathrm{at}\:\mathrm{time}\:\mathrm{t}\:=\:\mathrm{1}. \\$$

Answered by robocop last updated on 16/Feb/17

$${v}=\mathrm{18}+\mathrm{6}{xy}+\mathrm{3}+\mathrm{50} \\$$$${a}={v}' \\$$$${a}=\mathrm{6}\left({x}+{y}\right) \\$$

Commented by Saham last updated on 16/Feb/17

$$\mathrm{Thanks}\:\mathrm{sir}. \\$$

Commented by FilupS last updated on 18/Feb/17

$$\mathrm{incorrect}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{refering}\:\mathrm{to} \\$$$$\mathrm{linear}\:\mathrm{velocity} \\$$$$\mathrm{i}.\mathrm{e}.\:\:\:{v}\in\mathbb{R} \\$$$$\mathrm{This}\:\mathrm{question}\:\mathrm{is}\:\mathrm{about}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{field} \\$$$$\mathrm{i}.\mathrm{e}.\:{v}\in\mathbb{R}^{\mathrm{3}} \\$$

Answered by ajfour last updated on 16/Feb/17

$${a}_{{x}} =\mathrm{2}\left({yv}_{{x}} +{xv}_{{y}} +{t}\right);\:{at}\:{that}\:{time}\:{and}\:{place}\:{is}\:=\mathrm{2}\left[\mathrm{0}+\mathrm{3}\left(−\mathrm{10}\right)+\mathrm{1}\right]=−\mathrm{58} \\$$$${a}_{{y}} =−\left({y}^{\mathrm{2}} {v}_{{x}} +\mathrm{2}{xyv}_{{y}} +\mathrm{10}\right);\:{then}\:{and}\:{there}\:{will}\:{be}\:=−\left[\mathrm{0}+\mathrm{0}+\mathrm{10}\right]=−\mathrm{10} \\$$$${a}_{{z}} =\mathrm{0} \\$$$${So}\:\bar {{a}}\:{then}\:{and}\:{there}\:\left(\mathrm{3},\mathrm{0},\mathrm{2},\mathrm{1}\right)\:{would}\:{be}\:−\mathrm{58}\hat {{i}}−\mathrm{10}\hat {{j}}. \\$$

Commented by Saham last updated on 16/Feb/17

$$\mathrm{Thanks}\:\mathrm{sir}. \\$$

Answered by mrW1 last updated on 16/Feb/17

$${v}_{{x}} =\mathrm{6}+\mathrm{2}{xy}+{t}^{\mathrm{2}} \\$$$${v}_{{y}} =−\left({xy}^{\mathrm{2}} +\mathrm{10}{t}\right) \\$$$${v}_{{z}} =\mathrm{25} \\$$$${a}_{{x}} =\frac{{dv}_{{x}} }{{dt}}=\mathrm{2}{x}\frac{{dy}}{{dt}}+\mathrm{2}{y}\frac{{dx}}{{dt}}+\mathrm{2}{t}=\mathrm{2}\left({xv}_{{y}} +{yv}_{{x}} +{t}\right) \\$$$${a}_{{y}} =\frac{{dv}_{{y}} }{{dt}}=−\left({xy}\frac{{dy}}{{dt}}+{y}^{\mathrm{2}} \frac{{dx}}{{dt}}+\mathrm{10}\right)=−\left({xyv}_{{y}} +{y}^{\mathrm{2}} {v}_{{x}} +\mathrm{10}\right) \\$$$${a}_{{z}} =\mathrm{0} \\$$$$\\$$$${at}\:\left(\mathrm{3},\mathrm{0},\mathrm{2}\right)\:{and}\:{t}=\mathrm{1}: \\$$$${v}_{{x}} =\mathrm{6}+\mathrm{2}×\mathrm{3}×\mathrm{0}+\mathrm{1}^{\mathrm{2}} =\mathrm{7} \\$$$${v}_{{y}} =−\left(\mathrm{3}×\mathrm{0}^{\mathrm{2}} +\mathrm{10}×\mathrm{1}\right)=−\mathrm{10} \\$$$${v}_{{z}} =\mathrm{25} \\$$$${a}_{{x}} =\mathrm{2}\left[\mathrm{3}×\left(−\mathrm{10}\right)+\mathrm{0}×\mathrm{7}+\mathrm{1}\right]=−\mathrm{58} \\$$$${a}_{{y}} =−\left[\mathrm{3}×\mathrm{0}×\left(−\mathrm{10}\right)+\mathrm{0}^{\mathrm{2}} ×\mathrm{7}+\mathrm{10}\right]=−\mathrm{10} \\$$$${a}_{{z}} =\mathrm{0} \\$$$$\Rightarrow{a}=−\mathrm{58}{i}−\mathrm{10}{j}+\mathrm{0}{k} \\$$

Commented by Saham last updated on 16/Feb/17

$$\mathrm{Thanks}\:\mathrm{sir}. \\$$