A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $$f2$$ to come to rest. If the total distance traversed is $$15$$ S, then

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A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $$f2$$ to come to rest. If the total distance traversed is $$15$$ S, then

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Let car starts from point A from rest and moves up to point B with acceleration f.Velocity of car at point B, v $$=2fS$$                                                                [As $$v2=$$ $$u2+2as$$]x $$=$$ $$2fS$$ .t  −−−(i)        [As s $$=$$ $$ut}So$$ the velocity of car at point C also will be $$2fs$$ and finally car stops after covering distance y.Distance CD i.e, y $$=$$ (2fS)22(f2) $$=$$ $$2fSf$$ $$=$$ $$2S$$  ...(ii)                               [As $$v2=$$ $$u2$$−$$2as$$  s $$=$$ $$u2/2a$$]So, total distance AD $$=$$ AB $$+BC$$ $$+CD$$ $$=$$ $$15S(given)$$⇒ $$S+x+2S$$ $$=$$ $$15S$$  ⇒ x $$=$$ $$12SSubstituting$$ the value of x in equation (i) we getX $$=$$  $$2fS$$.t ⇒ $$12S$$ $$=$$ $$2fS$$ .t⇒ $$144S2$$ $$=$$ $$2fS$$.$$t2$$⇒ S $$=$$ $$172ft2$$

A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f2 to come to rest. If the total distance traversed is 15 S, then

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