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Question Number 136353 by BHOOPENDRA last updated on 21/Mar/21

Commented by BHOOPENDRA last updated on 21/Mar/21

mr.W sir here is full question ?

$${mr}.{W}\:{sir}\:{here}\:{is}\:{full}\:{question}\:? \\ $$

Answered by mr W last updated on 21/Mar/21

Commented by mr W last updated on 21/Mar/21

Part I  a(t)= { ((0 for 0≤t<t_1 )),((−c(t−t_1 ) for t_1 ≤t≤t_2 )) :}    0≤t<t_1 :  v(t)=v_0   s(t)=v_0 t  s(t_1 )=v_0 t_1     t_1 ≤t≤t_2 :  v(t)=v_0 +∫_t_1  ^t a(t)dt           =v_0 −c∫_t_1  ^t (t−t_1 )dt           =v_0 −c[(((t−t_1 )^2 )/2)]_t_1  ^t            =v_0 −((c(t−t_1 )^2 )/2)  v(t_2 )=v_0 −((c(t_2 −t_1 )^2 )/2)=0  ⇒v_0 =((c(t_2 −t_1 )^2 )/2)  s(t)=s(t_1 )+∫_t_1  ^t v(t)dt         =v_0 t_1 +∫_t_1  ^t [v_0 −((c(t−t_1 )^2 )/2)]dt         =v_0 t_1 +[v_0 t−((c(t−t_1 )^3 )/6)]_t_1  ^t          =v_0 t_1 +[v_0 t−((c(t−t_1 )^3 )/6)−v_0 t_1 ]         =v_0 t−((c(t−t_1 )^3 )/6)

$$\boldsymbol{{Part}}\:\boldsymbol{{I}} \\ $$$${a}\left({t}\right)=\begin{cases}{\mathrm{0}\:{for}\:\mathrm{0}\leqslant{t}<{t}_{\mathrm{1}} }\\{−{c}\left({t}−{t}_{\mathrm{1}} \right)\:{for}\:{t}_{\mathrm{1}} \leqslant{t}\leqslant{t}_{\mathrm{2}} }\end{cases} \\ $$$$ \\ $$$$\mathrm{0}\leqslant{t}<{t}_{\mathrm{1}} : \\ $$$${v}\left({t}\right)={v}_{\mathrm{0}} \\ $$$${s}\left({t}\right)={v}_{\mathrm{0}} {t} \\ $$$${s}\left({t}_{\mathrm{1}} \right)={v}_{\mathrm{0}} {t}_{\mathrm{1}} \\ $$$$ \\ $$$${t}_{\mathrm{1}} \leqslant{t}\leqslant{t}_{\mathrm{2}} : \\ $$$${v}\left({t}\right)={v}_{\mathrm{0}} +\int_{{t}_{\mathrm{1}} } ^{{t}} {a}\left({t}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:={v}_{\mathrm{0}} −{c}\int_{{t}_{\mathrm{1}} } ^{{t}} \left({t}−{t}_{\mathrm{1}} \right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:={v}_{\mathrm{0}} −{c}\left[\frac{\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}}\right]_{{t}_{\mathrm{1}} } ^{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:={v}_{\mathrm{0}} −\frac{{c}\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${v}\left({t}_{\mathrm{2}} \right)={v}_{\mathrm{0}} −\frac{{c}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\frac{{c}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${s}\left({t}\right)={s}\left({t}_{\mathrm{1}} \right)+\int_{{t}_{\mathrm{1}} } ^{{t}} {v}\left({t}\right){dt} \\ $$$$\:\:\:\:\:\:\:={v}_{\mathrm{0}} {t}_{\mathrm{1}} +\int_{{t}_{\mathrm{1}} } ^{{t}} \left[{v}_{\mathrm{0}} −\frac{{c}\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}}\right]{dt} \\ $$$$\:\:\:\:\:\:\:={v}_{\mathrm{0}} {t}_{\mathrm{1}} +\left[{v}_{\mathrm{0}} {t}−\frac{{c}\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{3}} }{\mathrm{6}}\right]_{{t}_{\mathrm{1}} } ^{{t}} \\ $$$$\:\:\:\:\:\:\:={v}_{\mathrm{0}} {t}_{\mathrm{1}} +\left[{v}_{\mathrm{0}} {t}−\frac{{c}\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{3}} }{\mathrm{6}}−{v}_{\mathrm{0}} {t}_{\mathrm{1}} \right] \\ $$$$\:\:\:\:\:\:\:={v}_{\mathrm{0}} {t}−\frac{{c}\left({t}−{t}_{\mathrm{1}} \right)^{\mathrm{3}} }{\mathrm{6}} \\ $$

Commented by mr W last updated on 21/Mar/21

Commented by BHOOPENDRA last updated on 21/Mar/21

thanku sir

$${thanku}\:{sir} \\ $$

Commented by BHOOPENDRA last updated on 21/Mar/21

part b sir?

$${part}\:{b}\:{sir}? \\ $$

Commented by mr W last updated on 21/Mar/21

Part II  given c=6 m/s^3 , t_1 =1 s, v_0 =12 m/s^2   v_0 =((c(t_2 −t_1 )^2 )/2)=((6(t_2 −1)^2 )/2)=12  ⇒t_2 =3 s  s(t_2 )=v_0 t_2 −((c(t_2 −t_1 )^3 )/6)      =12×3−((6(3−1)^3 )/6)=28 m  velocity of biker:  v_B =((17+28)/3)=15 m/s

$$\boldsymbol{{Part}}\:\boldsymbol{{II}} \\ $$$${given}\:{c}=\mathrm{6}\:{m}/{s}^{\mathrm{3}} ,\:{t}_{\mathrm{1}} =\mathrm{1}\:{s},\:{v}_{\mathrm{0}} =\mathrm{12}\:{m}/{s}^{\mathrm{2}} \\ $$$${v}_{\mathrm{0}} =\frac{{c}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{6}\left({t}_{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{12} \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\mathrm{3}\:{s} \\ $$$${s}\left({t}_{\mathrm{2}} \right)={v}_{\mathrm{0}} {t}_{\mathrm{2}} −\frac{{c}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\:\:\:\:=\mathrm{12}×\mathrm{3}−\frac{\mathrm{6}\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{6}}=\mathrm{28}\:{m} \\ $$$${velocity}\:{of}\:{biker}: \\ $$$${v}_{{B}} =\frac{\mathrm{17}+\mathrm{28}}{\mathrm{3}}=\mathrm{15}\:{m}/{s} \\ $$

Commented by BHOOPENDRA last updated on 21/Mar/21

thanks sir

$${thanks}\:{sir} \\ $$

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