Question

plz solve in detail
need urgent

Answer 1

Heya......!!!!


Nice Question :-))


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Let there be 2 points ,A and B and the Car starts from A from rest and moves with constant acceleration ( f ) to point B .

distance AB = S

we can use equation of motion here , to take out velocity of car at point B .

=> v^2 = u^2 + 2fS

u = 0

=> velocity of car at point B = √(2fS)


Now let there be another point C i.e from B-C and according to the question the car moves this distance ( BC ) with constant velocity

Let the distance BC = x ★

car moves at constant velocity in time t ,, so we can take out the distance ' x '

=> x = vt

here ,, v = √(2fS)

=> x = √(2fS) × t


Now let there be another point D i.e CD and according to the question the car moves up to point D with constant retardation.

constant retardation = f / 2 ( Given )

Using - v^2 = u^2 + 2as

We know that velocity of car at point C will also be √(2fS)

Let the distance CD = y

according to the question the car stops after covering the distance y .

• v = 0

• u = √(2fS)

• a = f / 2

• s = y


➡ 0^2 = {√(2fS) }^2 - 2× f/ 2 × y

=>> y = 2fS / f

=> y = 2S ★


Total Distance is given to us in the question

Total Distance => 15 S

AB + BC + CD

=> S + x + 2S = 15 S

=> x = 12 S ★

and as solved above :- x = √(2fS) × t => 12S

=> squaring both the sides

➡ 144S^2 = 2fS×t^2


♠ S = 1 / 72 × ft^2

option ( c )




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Hope It Helps You ☺