My Science Problems (Part-5)

Classical Mechanics Level 2

The trajectory of a particle in x − y x-y plane is given by-

x = 7 y + 2 y 2 x=7y+2y^2

The acceleration of particle is constant and having magnitude 8 ms − 2 8 \text{ ms}^{-2} directed in positive x x direction.

Find the velocity of particle at origin (in ms − 1 \text{ms}^{-1} ).


The answer is 10.

Akshay Yadav by Akshay Yadav , 20, India

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1 solution

Rishabh Jain
May 13, 2016

I did everything by scratch!!

Differentiate the eqn of trajectory to get: v x = 7 v y + 4 y v y ( ∵ d x d t = v x , d y d t = v y ) \large v_x=7v_y+4yv_y~~(\color{#D61F06}{\small{\because\frac{dx}{dt}=v_x,\frac{dy}{dt}=v_y}}) Differentiate again: a x = 7 a y + 4 ( v y ) 2 + 4 y a y \large a_x=7a_y+4(v_y)^2+4ya_y ( ∵ d v x d t = a x , d v y d t = a y ) (\color{#D61F06}{\small{\because\frac{dv_x}{dt}=a_x,\frac{dv_y}{dt}=a_y}})

Acc to ques, a x = 8 ms − 2 , a y = 0 , x = y = 0 a_x=8\text{ ms}^{-2},a_y=0,x=y=0 , therefore direct substitution in formed eqns give: v x = 7 ms − 1 2 , v y = 2 ms − 1 \large v_x=7\text{ms}^{-1}\sqrt 2,v_y=\sqrt 2\text{ms}^{-1} ⟹ v = ( 7 2 ) 2 + ( 2 ) 2 = 10 ms − 1 \large \implies v=\sqrt{(7\sqrt 2)^2+(\sqrt 2)^2}=\Large\boxed{10\text{ms}^{-1}}

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What should be the level to this question according to you?

Akshay Yadav - 5 years, 1 month ago

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I don't have better knowledge of classical mechanics on brilliant maybe it'd be Level 3 − 4 3-4 .

Rishabh Jain - 5 years, 1 month ago

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