Question Number 14215 by tawa tawa last updated on 29/May/17
Answered by ajfour last updated on 29/May/17
![•(5). x=12t^2 −2t^3 =2t^2 (6−t) ....(i) at t=3s , x=18×3=54m v=(dx/dt)=24t−6t^2 =6t(4−t) ...(ii) at t=3s, v=18(4−3)=18m/s a=(dv/dt)=24−12t=12(2−t) ...(iii) at t=3s, a=12(2−3)=−12m/s^2 x is maximum when v=0, i.e. at t=4s [see (ii)] x_(max) =2t^2 (6−t)∣_(t=4s) =2(16)(2)=64m v is maximum when a=0 which occurs at t=2s [see (iii)] v_(max) =6t(4−t)∣_(t=2s) =12(2)=24m/s^2 the acceleration when it is not moving other than at t=0, is a at t=4s a=12(2−t)∣_(t=4s) = −24m/s^2 average velocity between t=0 and t=3s is v^ =((x_3 −x_0 )/(3−0))=((54m)/(3s))=18m/s. •(6). let that the vehicle accelerates for time t_1 . v=u+at ⇒ 20=0+2t_1 t_1 =10s it starts deceleeating thereafter and comes to a stop . let time taken to stop since t=10s, be t_2 : v=u+at ⇒ 0=20−1×t_2 t_2 =20s. Entire time taken from rest to stop is = t_1 +t_2 =10s+20s=30s. total distance travelled : s=s_1 +s_2 from s=ut+(1/2)at^2 s=[0+(1/2)(2)(10)^2 ]+[20×20−(1/2)(1)(20)^2 ] =100+200 =300m .](https://www.tinkutara.com/question/Q14220.png)
$$\bullet\left(\mathrm{5}\right). \\ $$$${x}=\mathrm{12}{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} =\mathrm{2}{t}^{\mathrm{2}} \left(\mathrm{6}−{t}\right)\:\:\:\:\:….\left({i}\right) \\ $$$$ \\ $$$${at}\:{t}=\mathrm{3}{s}\:,\:\:{x}=\mathrm{18}×\mathrm{3}=\mathrm{54}{m} \\ $$$${v}=\frac{{dx}}{{dt}}=\mathrm{24}{t}−\mathrm{6}{t}^{\mathrm{2}} =\mathrm{6}{t}\left(\mathrm{4}−{t}\right)\:\:\:…\left({ii}\right) \\ $$$${at}\:{t}=\mathrm{3}{s},\:\:{v}=\mathrm{18}\left(\mathrm{4}−\mathrm{3}\right)=\mathrm{18}{m}/{s} \\ $$$${a}=\frac{{dv}}{{dt}}=\mathrm{24}−\mathrm{12}{t}=\mathrm{12}\left(\mathrm{2}−{t}\right)\:\:\:…\left({iii}\right) \\ $$$${at}\:{t}=\mathrm{3}{s},\:\:{a}=\mathrm{12}\left(\mathrm{2}−\mathrm{3}\right)=−\mathrm{12}{m}/{s}^{\mathrm{2}} \\ $$$${x}\:{is}\:{maximum}\:{when}\:{v}=\mathrm{0},\:{i}.{e}. \\ $$$${at}\:{t}=\mathrm{4}{s}\:\:\:\:\:\:\:\left[{see}\:\left({ii}\right)\right] \\ $$$${x}_{{max}} =\mathrm{2}{t}^{\mathrm{2}} \left(\mathrm{6}−{t}\right)\mid_{{t}=\mathrm{4}{s}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{16}\right)\left(\mathrm{2}\right)=\mathrm{64}{m} \\ $$$${v}\:{is}\:{maximum}\:{when}\:{a}=\mathrm{0} \\ $$$${which}\:{occurs}\:{at}\:{t}=\mathrm{2}{s}\:\:\left[{see}\:\left({iii}\right)\right] \\ $$$${v}_{{max}} =\mathrm{6}{t}\left(\mathrm{4}−{t}\right)\mid_{{t}=\mathrm{2}{s}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{12}\left(\mathrm{2}\right)=\mathrm{24}{m}/{s}^{\mathrm{2}} \\ $$$${the}\:{acceleration}\:{when}\:{it}\:{is} \\ $$$${not}\:{moving}\:{other}\:{than}\:{at}\:{t}=\mathrm{0}, \\ $$$${is}\:\boldsymbol{{a}}\:{at}\:{t}=\mathrm{4}{s} \\ $$$${a}=\mathrm{12}\left(\mathrm{2}−{t}\right)\mid_{{t}=\mathrm{4}{s}} =\:−\mathrm{24}{m}/{s}^{\mathrm{2}} \\ $$$${average}\:{velocity}\:{between} \\ $$$${t}=\mathrm{0}\:{and}\:{t}=\mathrm{3}{s}\:{is} \\ $$$$\bar {{v}}=\frac{{x}_{\mathrm{3}} −{x}_{\mathrm{0}} }{\mathrm{3}−\mathrm{0}}=\frac{\mathrm{54}{m}}{\mathrm{3}{s}}=\mathrm{18}{m}/{s}. \\ $$$$\bullet\left(\mathrm{6}\right). \\ $$$${let}\:{that}\:{the}\:{vehicle}\:{accelerates} \\ $$$${for}\:{time}\:\boldsymbol{{t}}_{\mathrm{1}} . \\ $$$$\boldsymbol{{v}}=\boldsymbol{{u}}+\boldsymbol{{at}}\:\:\Rightarrow\:\:\:\mathrm{20}=\mathrm{0}+\mathrm{2}{t}_{\mathrm{1}} \\ $$$${t}_{\mathrm{1}} =\mathrm{10}{s} \\ $$$${it}\:{starts}\:{deceleeating}\:{thereafter} \\ $$$${and}\:{comes}\:{to}\:{a}\:{stop}\:. \\ $$$${let}\:\:{time}\:{taken}\:{to}\:{stop}\:{since} \\ $$$${t}=\mathrm{10}{s},\:{be}\:{t}_{\mathrm{2}} : \\ $$$$\boldsymbol{{v}}=\boldsymbol{{u}}+\boldsymbol{{at}}\:\:\Rightarrow\:\:\mathrm{0}=\mathrm{20}−\mathrm{1}×{t}_{\mathrm{2}} \\ $$$${t}_{\mathrm{2}} =\mathrm{20}{s}. \\ $$$${Entire}\:{time}\:{taken}\:{from}\:{rest} \\ $$$${to}\:{stop}\:{is}\:\:=\:{t}_{\mathrm{1}} +{t}_{\mathrm{2}} =\mathrm{10}{s}+\mathrm{20}{s}=\mathrm{30}{s}. \\ $$$${total}\:{distance}\:{travelled}\:: \\ $$$${s}={s}_{\mathrm{1}} +{s}_{\mathrm{2}} \\ $$$$\:\:{from}\:\boldsymbol{{s}}=\boldsymbol{{ut}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{at}}^{\mathrm{2}} \\ $$$$\boldsymbol{{s}}=\left[\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{10}\right)^{\mathrm{2}} \right]+\left[\mathrm{20}×\mathrm{20}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{20}\right)^{\mathrm{2}} \right] \\ $$$$\:\:=\mathrm{100}+\mathrm{200}\:=\mathrm{300}\boldsymbol{{m}}\:. \\ $$
Commented by tawa tawa last updated on 29/May/17
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{i}\:\mathrm{must}\:\mathrm{have}\:\mathrm{an}\:\mathrm{As}\:\mathrm{in}\:\mathrm{my}\:\mathrm{mathematics}\:\mathrm{and}\:\mathrm{physics}. \\ $$
Commented by ajfour last updated on 29/May/17
Commented by ajfour last updated on 29/May/17
$${for}\:\:\bullet\left(\mathrm{6}\right). \\ $$