Question Number 43354 by pieroo last updated on 10/Sep/18
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{acceleration}\left(\mathrm{30}+\mathrm{6t}\right) \\ $$$$\mathrm{ms}^{−\mathrm{2}} \:\mathrm{at}\:\mathrm{time}\:\mathrm{t}.\:\mathrm{Where}\:\mathrm{will}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{come}\:\mathrm{to}\:\mathrm{rest} \\ $$$$\mathrm{again}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
$$\frac{{dv}}{{dt}}=\mathrm{30}+\mathrm{6}{t} \\ $$$${here}\:{accelaration}\:{is}\:+{ve}\:{so}\:{no}\:{question} \\ $$$${arise}\:{that}\:{partcle}\:{will}\:{come}\:{to}\:{rest}\:{again} \\ $$$${dv}=\left(\mathrm{30}+\mathrm{6}{t}\right){dt} \\ $$$${v}=\mathrm{30}{t}+\mathrm{3}{t}^{\mathrm{2}} +{c} \\ $$$${t}=\mathrm{0}\:\:\:\:{v}=\mathrm{0}\:\:\:\:\:{so}\:{c}=\mathrm{0} \\ $$$${v}=\mathrm{30}{t}+\mathrm{3}{t}^{\mathrm{2}} \:\:{if}\:{particle}\:{at}\:{rest}\:\:{v}=\mathrm{0} \\ $$$$\mathrm{30}{t}+\mathrm{3}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}{t}\left(\mathrm{10}+{t}\right)=\mathrm{0} \\ $$$${t}=\mathrm{0}\:\:{but}\:{t}\:{can}\:{not}\:{be}\:{negetive}\:\left({t}=−\mathrm{10}\right) \\ $$$${so}\:{particle}\:{can}\:{never}\:{come}\:{to}\:{rest}\:{again}... \\ $$$$ \\ $$$$ \\ $$$$ \\ $$