Question Number 47815 by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
$${basic}\:{question} \\ $$$$\left.\mathrm{1}\right){fig}\:{a}\:\:{is}\:{it}\:{accelarated}\:{or}\:{deaccelarated}\:{motion} \\ $$$$\left.\mathrm{2}\right){fig}\:{b}\:{accelarated}/{deaccelarated}\:{motion} \\ $$$$\left.\mathrm{3}\right){fig}\left({a}\right)\:{particle}\:{moving}\:{towards}\:{origin}\:{or}\:{away} \\ $$$$\left.\mathrm{4}\right){fig}\:{b}\:{particle}\:{moving}\:{towards}\:{origin}\:{or}\:{away} \\ $$$${questions}\:{for}\:\mathrm{10}+\mathrm{2}\:{students}… \\ $$
Commented by mr W last updated on 15/Nov/18
$$\left.\mathrm{1}\right)\:{deaccelarated}\:{motion} \\ $$$$\left.\mathrm{2}\right)\:{deaccelarated}\:{motion} \\ $$$$\left.\mathrm{3}\right)\:{away}\:{from}\:{origin} \\ $$$$\left.\mathrm{4}\right)\:{towards}\:{origin} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Nov/18
$${answer}\:{with}\:{explanation}… \\ $$$${fig}\left({a}\right) \\ $$$${slope}\:{at}\:{t}_{\mathrm{1}} >{slope}\:{at}\:{t}_{\mathrm{2}} \\ $$$$\left({velocity}\right)_{{t}_{\mathrm{1}} } >\left({velocity}\right)_{{t}_{\mathrm{2}} } \:{so}\:{velocity}\:{decreases} \\ $$$${as}\:{time}\:{increases}\:{hence}\:{deaccelaration} \\ $$$${now}\:{bothe}\:{slope}\:{for}\:{acute}\:{angle}\:{so}\:{particle} \\ $$$${moving}\:{away}\:{from}\:{origin}. \\ $$$${fig}\left({b}\right) \\ $$$$\mid{tan}\left(\theta\right)_{{t}_{\mathrm{1}} } \mid>\mid{tan}\left(\theta\right)_{{t}_{\mathrm{2}} } \mid \\ $$$${so}\:{velocity}\:{decreases}\:{hence}\:{deaccelaration} \\ $$$${but}\:{in}\:{fig}\left({b}\right)\:{angle}\:{obtuse}\:{so}\:{particle}\:{moving} \\ $$$${towards}\:{origin}… \\ $$