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# The position of a body in one dimensional motion varies with time as x=24t -3t2 . What is the nature of its motion? Find a) its velocity after 4 sec. b) Its acceleration and c) its average velocity from t=0 during the interval when the velocity is positive.

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## Given, $\mathrm{x}=24\mathrm{t}-3{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=24-6\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{body}\mathrm{is}\mathrm{moving}\mathrm{with}\mathrm{uniform}\mathrm{acceleration}.\phantom{\rule{0ex}{0ex}}\left(\mathrm{a}\right)\mathrm{Velocity}\mathrm{after}4\mathrm{sec}\phantom{\rule{0ex}{0ex}}\mathrm{v}=24-6×4=0\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)\mathrm{Acceleration}=-6\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)\mathrm{Since}\mathrm{velocity}\mathrm{after}\mathrm{t}=1\mathrm{sec}\phantom{\rule{0ex}{0ex}}\mathrm{v}=24-6×1=18\mathrm{m}/\mathrm{s}\left(\mathrm{positive}\mathrm{value}\right)\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{t}=0\phantom{\rule{0ex}{0ex}}\mathrm{Position},\mathrm{x}=24×0-3×0=0\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{t}=1\phantom{\rule{0ex}{0ex}}\mathrm{x}\text{'}=24×1-3×{1}^{2}=21\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Displacement}\mathrm{in}\mathrm{this}\mathrm{time}\mathrm{interval},\phantom{\rule{0ex}{0ex}}\mathrm{x}\text{'}-\mathrm{x}=21-0=21\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Average}\mathrm{velocity}=\frac{\mathrm{Displacement}}{\mathrm{Time}\mathrm{interval}}=\frac{21}{\left(1-0\right)}=21\mathrm{m}/\mathrm{s}$  Suggest Corrections  0      Similar questions
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