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Question Number 126092 by physicstutes last updated on 17/Dec/20

$$\mathrm{A}\mathrm{particle}\mathrm{starts}\mathrm{from}\mathrm{rest}\mathrm{at}\mathrm{time}t=0\mathrm{and}\mathrm{moves}\mathrm{in}$$$$\mathrm{a}\mathrm{straightline}\mathrm{with}\mathrm{variable}\mathrm{acceleration}a\mathrm{m}/{\mathrm{s}}^2\mathrm{where}$$$$a=\frac{t}{5},0⩽t⩽5,a=\frac{t}{5}+\frac{10}{t^2},t⩾5,t\mathrm{being}\mathrm{measured}\mathrm{in}\mathrm{seconds}.$$$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{velocity}\mathrm{is}22\frac{1}{2}\mathrm{m}/\mathrm{s}\mathrm{when}t=5\mathrm{and}$$$$11\mathrm{m}/\mathrm{s}\mathrm{when}t=10.$$$$\mathrm{Show}\mathrm{also}\mathrm{that}\mathrm{the}\mathrm{distance}\mathrm{travelled}\mathrm{by}\mathrm{the}\mathrm{particle}$$$$\mathrm{in}\mathrm{the}\mathrm{first}10\mathrm{seconds}\mathrm{is}(43\frac{1}{3}-10\ln 2)\mathrm{m}.$$

Answered by ajfour last updated on 17/Dec/20

$$v={\int }_0^t\frac{t}{5}dt;0⩽t⩽5$$$$=\frac{5}{2}+{\int }_5^t(\frac{t}{5}+\frac{10}{t^2})dt;t>5$$$$=\frac{5}{2}+(\frac{t^2}{10}-\frac{10}{t})-(\frac{5}{2}-2)$$$$\Rightarrow Att=5,v=2\frac{1}{2}m/s=\frac{5}{2}m/s,$$$$andatt=10,v=11m/s.$$$$Distancesuntilt=10is{s}_1+s_2,\mathrm{\forall }$$$$s_1={\int }_0^5\left(\frac{t^2}{10}\right)dt=\frac{125}{30}=\frac{25}{6}$$$$s_2={\int }_5^{10}(\frac{t^2}{10}-\frac{10}{t}+2)dt$$$$=(\frac{t^3}{30}-10\ln t+2t){\mid }_5^{10}$$$$=(\frac{100}{3}-\frac{25}{6}-10\ln 2+10)$$$$s_1+s_2=43\frac{1}{3}-10\ln 2.(Indeed!)$$

Answered by Dwaipayan Shikari last updated on 17/Dec/20

$$\stackrel{.}{x}=\int \stackrel{..}{xdt}={\int }_0^5\frac{t}{5}dt=\frac{25}{10}.\frac{m}{s}=2.5\frac{m}{s}$$$$After10seconds$$$$\stackrel{.}{x}=2.5+{\int }_5^{10}\frac{t}{5}+\frac{10}{t^2}dt=7.5+1+2.5\frac{m}{s}=11\frac{m}{s}$$

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