Menu Close

Question-57084

Tinku Tara 0 Comments
Pin




Question Number 57084 by frimpshaddie last updated on 30/Mar/19
Commented by Kunal12588 last updated on 30/Mar/19
(d) when the velocity of particle becomes 0 then the particle is said to be in rest (horizontal motion only) ∴v=0 ⇒3t^2 −6t=0 ⇒3t(t−2)=0 ⇒t=0 or t=2 at t=0s s=(0)^3 −2(0)^2 +1=1m at t=2s s=(2)^3 −2(2)^2 +1=8−8+1=1m when :− time is equal to 0s or 2s where :− distance is equal to 1m
$$\left({d}\right)\:{when}\:{the}\:{velocity}\:{of}\:{particle}\:{becomes}\:\mathrm{0} \\ $$$${then}\:{the}\:{particle}\:{is}\:{said}\:{to}\:{be}\:{in}\:{rest} \\ $$$$\left({horizontal}\:{motion}\:{only}\right) \\ $$$$\therefore{v}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{t}\left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{0}\:{or}\:{t}=\mathrm{2} \\ $$$${at}\:{t}=\mathrm{0}{s} \\ $$$${s}=\left(\mathrm{0}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{1}{m} \\ $$$${at}\:{t}=\mathrm{2}{s} \\ $$$${s}=\left(\mathrm{2}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{8}−\mathrm{8}+\mathrm{1}=\mathrm{1}{m} \\ $$$${when}\::−\:{time}\:{is}\:{equal}\:{to}\:\mathrm{0}{s}\:{or}\:\mathrm{2}{s} \\ $$$${where}\::−\:{distance}\:{is}\:{equal}\:{to}\:\mathrm{1}{m} \\ $$
Commented by kkc last updated on 30/Mar/19
$$ \\ $$
Answered by Kunal12588 last updated on 30/Mar/19
(a) v=(ds/dt)=(d/dt)(t^3 −2t^2 +1) ⇒v=3t^2 −6t t=2s ⇒v=3(2)^2 −6(2)= 0 ms^(−1)
$$\left({a}\right)\:{v}=\frac{{ds}}{{dt}}=\frac{{d}}{{dt}}\left({t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow{v}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t} \\ $$$${t}=\mathrm{2}{s} \\ $$$$\Rightarrow{v}=\mathrm{3}\left(\mathrm{2}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{2}\right)=\:\mathrm{0}\:{ms}^{−\mathrm{1}} \\ $$
Answered by Kunal12588 last updated on 30/Mar/19
(b) v_(av) =((s_(final) −s_(initial) )/(t_(final) −t_(initial) )) t_(initial) =0s s_(initial) =(0)^3 −2(0)^2 +1=1 m t_(final) =3s s_(final) =(3)^3 −2(3)^2 +1=27−18+1=10m ∴v_(av) =((10−1)/(3−0))=(9/3)=3ms^(−1)
$$\left({b}\right)\:{v}_{{av}} =\frac{{s}_{{final}} −{s}_{{initial}} }{{t}_{{final}} −{t}_{{initial}} } \\ $$$${t}_{{initial}} =\mathrm{0}{s} \\ $$$${s}_{{initial}} =\left(\mathrm{0}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{1}\:{m} \\ $$$${t}_{{final}} =\mathrm{3}{s} \\ $$$${s}_{{final}} =\left(\mathrm{3}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{27}−\mathrm{18}+\mathrm{1}=\mathrm{10}{m} \\ $$$$\therefore{v}_{{av}} =\frac{\mathrm{10}−\mathrm{1}}{\mathrm{3}−\mathrm{0}}=\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{3}{ms}^{−\mathrm{1}} \\ $$
Answered by Kunal12588 last updated on 30/Mar/19
(c)a_(av) =((v_2 −v_1 )/(t_2 −t_1 )) v=3t^2 −6t t_2 =3s ⇒v_2 =3(3)^2 −6(3)=27−18=9 ms^(−1) t_1 =1s v_1 =3(1)^2 −6(1)=3−6=−3ms^(−1) ∴a_(av) =((9−(−3))/(3−1))=((12)/2)=6ms^(−2)
$$\left({c}\right){a}_{{av}} =\frac{{v}_{\mathrm{2}} −{v}_{\mathrm{1}} }{{t}_{\mathrm{2}} −{t}_{\mathrm{1}} } \\ $$$${v}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t} \\ $$$${t}_{\mathrm{2}} =\mathrm{3}{s} \\ $$$$\Rightarrow{v}_{\mathrm{2}} =\mathrm{3}\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{3}\right)=\mathrm{27}−\mathrm{18}=\mathrm{9}\:{ms}^{−\mathrm{1}} \\ $$$${t}_{\mathrm{1}} =\mathrm{1}{s} \\ $$$${v}_{\mathrm{1}} =\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{1}\right)=\mathrm{3}−\mathrm{6}=−\mathrm{3}{ms}^{−\mathrm{1}} \\ $$$$\therefore{a}_{{av}} =\frac{\mathrm{9}−\left(−\mathrm{3}\right)}{\mathrm{3}−\mathrm{1}}=\frac{\mathrm{12}}{\mathrm{2}}=\mathrm{6}{ms}^{−\mathrm{2}} \\ $$
Answered by kkc last updated on 30/Mar/19
$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *