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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
S=12ft2
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B
S=14ft2
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C
S=172ft2
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D
S=16ft2
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Solution

The correct option is C S=172ft2
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=u22as 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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