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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 f2 to come to rest. If the total distance traversed is 15 S, then

A
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Solution

The correct option is C

Let car starts from point A from rest and moves up to point B with acceleration f

Car moves distance BC with this constant velocity in time t [AS v2=u2+2as]

x=2f St .....(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 Y=(2fS)22(f/2)=2fSf=2S ...(ii) [As v2=u22as s=u2/2a]

so, The total distance AD=AB+BC+CD=15 S(given)

S+x+2S=15S x=12S

Substituting the value of x in equation (i) we get


x=2fSt 12S=2fSt 144S2=2 fS.t2

s=172 ft2.


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