AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, initial speed of car, vi=v
Final speed of car when it stops, vf=0
Distance travelled =s
Retarding force =F
By Newton's law of motion,
F=ma=m(

vi2−vi2

2s

) ...(i) [

using‌relation

a= vf2−vi2 2s

]
=

m(−v2)

2s

F=

−mv2

2s

When, retarding force, F′=3F
Let s′= New distance travelled.
Similarly, from Eq. (i) we can write
3F=

−mv2

2s′

...(ii)
Substituting value of F from Eq. (i) into Eq. (ii), we get
3(

−mv2

2s

)=

−mv2

2s2

⇒ s′=

s

3

Hence, the body will stop by travelling distance

s

3

, when retarding force is 3F.