Question

Can anyone help me please......​

Answer 1

Solution :-

Part I :

• Initial velocity of the object = V•
• Final velocity of the object = V (max)

• Acceleration of the object = a
• Distance traveled by the object = L 1

As the acceleration remains constant throughout the first half of it's motion we can use the third kinematic equation in order to solve this question

$\implies\mathtt{ {v}^{2} = {u}^{2} + 2as}$

here ,

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance

Now let us substitute the given values in the above equation ,

$\implies\mathtt{{V} ^{2}= {V_o} ^{2}+ 2aL_1 }$

$\implies\mathtt{ {V} ^{2}-{V_o} ^{2} = 2aL_1 }$

$\implies\mathtt{L_1 = \dfrac{{V} ^{2}- {V_o} ^{2}}{ 2a} } ...(1)$

________________

Part II :-

The final velocity in Part I will become the initial velocity of the object in Part II

• Initial velocity = V (max)
• Final velocity = 0 ( as the car comes to rest )
• Acceleration = - a ( negative sign denotes retardation )
• Distance covered = L 2

By using the third equation of motion ,

$\implies\mathtt{ {v}^{2} = {u}^{2} + 2as}$

Now let us substitute the given values in the above equation ,

$\implies\mathtt{0 = {V} ^{2} - 2aL_2 }$

$\implies\mathtt{ {V} ^{2} = 2aL_2 }$

$\implies\mathtt{L_2= \dfrac{ {V} ^{2}}{ 2a} } ..(2)$

_____________

Now as per the given question ,

$\implies \mathtt{k = \dfrac{L_2}{L_1} }..(3)$

Substituting the values of (1) and (2) in (3) we get ,

$\implies\mathtt{ k = \dfrac{\dfrac{{V} ^{2}}{\cancel{2a}}}{\dfrac{{V} ^{2}- {V_o} ^{2}}{\cancel{2a}}}}$

$\implies\mathtt{k = \dfrac{ {V}^{2} }{ {V}^{2} - {V_o}^{2} }}$

$\implies\mathtt{k {V}^{2} - k {V_o}^{2} ={V} ^{2} }$

$\implies\mathtt{k {V}^{2} - {V}^{2} =k{V_o} ^{2} }$

$\implies\mathtt{{V}^{2} (k- 1) =k{V_o} ^{2} }$

$\implies\mathtt{ {V}^{2} =\dfrac{k} {k-1}{V_o} ^{2}}$

$\implies\mathtt{ \red{V = \sqrt{\dfrac{k} {k-1}} V_o}}$

Answer 2

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But enikk enganeyan find cheyya ennaryilla

enthayalum nokkam