Question Number 27117 by Tinkutara last updated on 02/Jan/18
Answered by ajfour last updated on 02/Jan/18
$$\frac{{dv}}{{dt}}=−\left(\frac{{v}^{\mathrm{2}} }{{r}}\right)\:,\:\:\left({becoz}\:\:{a}_{{t}} =−{a}_{{r}} =−\frac{{v}^{\mathrm{2}} }{{r}}\right) \\ $$$$\int_{\mathrm{10}} ^{\:\:\mathrm{5}} \frac{{dv}}{{v}^{\mathrm{2}} }=−\frac{\mathrm{1}}{{r}}\int_{{t}_{\mathrm{0}} } ^{\:\:{t}_{\mathrm{0}} +\bigtriangleup{t}} {dt} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{10}}=\frac{\bigtriangleup{t}}{\mathrm{20}} \\ $$$${or}\:\:\:\bigtriangleup{t}\:=\:\mathrm{2}{s}\:. \\ $$
Commented by Tinkutara last updated on 03/Jan/18
Thank you very much Sir! I got the answer.
Commented by ajfour last updated on 02/Jan/18
$${acceleration}\:{is}\:{directed}\:{at}\:\mathrm{45}°\:{to} \\ $$$${radius}\:,\:{tagential}\:{acc}.\:{and} \\ $$$${radial}\:{acc}.\:{are}\:{equal}\:{in}\:{magnitude} \\ $$$${then}.\:{But}\:{as}\:{tangential}\:{acc}.\:{turns} \\ $$$${out}\:{to}\:{be}\:{oppositely}\:{directed}\:{to} \\ $$$${velocity}\:{we}\:{take}\:\:{a}_{{t}} =\frac{{dv}}{{dt}}=−\frac{{v}^{\mathrm{2}} }{{r}}. \\ $$$${v}\:{is}\:{the}\:{speed}\:{of}\:{particle}\:{at}\:{any} \\ $$$${time}. \\ $$