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A car travels starting from rest with constant acceleration alpha and reaches a maximum velocity V.It travels with maximum velocity for some time and retards uniformly at the rate of beta and comes to rest.If s is the total distance and t is the total time of journey then t= |
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Answer» Solution :` ` `v=0+alpha1,,t_(1)(V)/(alpha)` `S_(1)=(1)/(2)alphat_(1)^(2)=(v^(2))/(2alpha)` O-V=`betat_(3),t_(3)=(v)/(BETA)` `O^(2)-V^(2)=-2|` `S_(3)=(v_(2))/(2beta)=(1)/(2)betat_(3)^(2)` `t_(2)=(S_(2))/(V)=(S-(S_(1)+S_(3)))/(V)=(S-(v^(2))/(2)((1)/(alpha)+(1)/(beta)))/(V)` `t=t_(1)+t_(2)+t_(3)` `=(v)/(alpha)+(s)/(v)-(v)/(2alpha)-(v)/(2beta)+(v)/(beta) t=(s)/(v)+(v)/(2)((alpha+beta)/(ALPHABETA))` `S_(2)=S-(S_(1)+S_(2))=S-((v^(2))/(2alpha)+(v^(2))/(2alpha))` `=S-(v^(2))/(alpha)`=s-(vt-s) `S_(2)`=2s-vt [`S_(2)`=distance TRAVELLED with constant velocity] |
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