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Question

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 f/2 to come at rest. If the total distance traversed is 15S, then-

A
S=12ft2
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B
S=14ft2
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C
S=ft
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D
S=172ft2
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Solution

The correct option is D S=172ft2
Distance traversed is S then velocity attained will be v=2fS ....... (1)
so the next distance covered with this constant velocity will be x=vt=2fS×t
now distance covered in the time when it comes to rest with retardation f will be y=v22(f/2)=v2f=2fSf=2S
(see equation-1)
net distance is S+x+y=S+t2fS+2S=3S+t2fS
given that the net distance is 15S so 15S=3S+t2fS
or (12S)2=t2×2fS so S=172ft2

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