Question

# A car, starting from rest, accelerates at the rate f through a distance S, then continuous 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. Velocity of car at point B, v=√2fs [As v2=u2+2as] Car moves distance BC with this constant velocity in time t. 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⇒ y=(√2fs)22(f2)=2fsf=2S ...(ii) [As v2=u2−2as⇒ s=u22a] 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=√2fs.t ⇒ 12S=√2fs.t ⇒ 144S2=2fs.t2 ⇒ S=172ft2.

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