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Question-47817




Question Number 47817 by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
same questions as asked in Q.No−47815  fig(a)accelarated/deaccelarated  fig(b)accelarated/deaccelarated  fig(a)particle moving twards origin/or away  fig(b)particle movinv towards origin/away
$${same}\:{questions}\:{as}\:{asked}\:{in}\:{Q}.{No}−\mathrm{47815} \\ $$$${fig}\left({a}\right){accelarated}/{deaccelarated} \\ $$$${fig}\left({b}\right){accelarated}/{deaccelarated} \\ $$$${fig}\left({a}\right){particle}\:{moving}\:{twards}\:{origin}/{or}\:{away} \\ $$$${fig}\left({b}\right){particle}\:{movinv}\:{towards}\:{origin}/{away} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
fig(a) both slope acute angle time t_1 and t_(2 ) so   particle moving away origin.  (slope)_t_2  >(slope)_t_1   so velocity increases  hence accelarated motion.  fig(b)  slope at t_1 and t_2  obtuse angle  so particle moving towsrds origin.  ∣tan(θ)_t_2  ∣>∣tan(θ)_t_1  ∣  so velocity increses  hence accelarated motion.
$${fig}\left({a}\right)\:{both}\:{slope}\:{acute}\:{angle}\:{time}\:{t}_{\mathrm{1}} {and}\:{t}_{\mathrm{2}\:} {so}\: \\ $$$${particle}\:{moving}\:{away}\:{origin}. \\ $$$$\left({slope}\right)_{{t}_{\mathrm{2}} } >\left({slope}\right)_{{t}_{\mathrm{1}} } \:{so}\:{velocity}\:{increases} \\ $$$${hence}\:{accelarated}\:{motion}. \\ $$$${fig}\left({b}\right) \\ $$$${slope}\:{at}\:{t}_{\mathrm{1}} {and}\:{t}_{\mathrm{2}} \:{obtuse}\:{angle} \\ $$$${so}\:{particle}\:{moving}\:{towsrds}\:{origin}. \\ $$$$\mid{tan}\left(\theta\right)_{{t}_{\mathrm{2}} } \mid>\mid{tan}\left(\theta\right)_{{t}_{\mathrm{1}} } \mid\:\:{so}\:{velocity}\:{increses} \\ $$$${hence}\:{accelarated}\:{motion}. \\ $$$$ \\ $$

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