Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com