Question

A snail starts its motion from origin at t = 0 and moves with constant acceleration any
direction in a X-Y plane. If its equation of motion is y= bx, then X-component of velocity is​

Answer 1

Given : A snail starts its motion from origin at t = 0 and moves with constant acceleration any  direction in a X-Y plane  

equation of motion is y= bx

To Find : X-component of velocity  

Solution:

y = bx

=> dy/dt = b dx/dt

V = √(dy/dt)² + (dx/dt)²

=> V = √(bdx/dt)² + (dx/dt)²

=>   V =√ (b² + 1)√( dx/dt)²

=> V = √(b² + 1) dx/dt

=> dV/dt  =  √ (b² + 1) d²x/dt²

a = dV/dt  = constant acceleration

=> a =  √(b² + 1) d²x/dt²

=>  d²x/dt²  = a/√(b² + 1)

integrating both sides

=> dx/dt = at/√(b² + 1)  + C

dx/dt  = 0 at  t = 0  as start from rest

=> 0 = 0 + C

=> C = 0

=> dx/dt = at/√(b² + 1)  + 0

dx/dt  = at/√(b² + 1) = X-component of velocity

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Answer 2

Given:

A snail starts its motion from origin at t = 0 and moves with constant acceleration any direction in a X-Y plane. Its equation of motion is y= bx.

To find:

x component of velocity ?

Calculation:

Let net acceleration be constant such that :

$\therefore \: \dfrac{dv}{dt} = k \: \: \: ...........(let)$

$\implies \: dv = k \: dt$

$\displaystyle \implies \: \int_{0}^{v} dv = k \: \int_{0}^{t} dt$

$\implies \: v = kt$

Now , as per equation of motion:

$\therefore \: y = bx$

$\implies \: \dfrac{dy}{dt} = b \times \dfrac{dx}{dt}$

$\implies \: v_{y}= b \times v_{x}$

Now , we know that net velocity is vector sum of the components :

$\therefore \: v = \sqrt{ {(v_{x})}^{2} + {(v_{y})}^{2} }$

$\implies \: v = \sqrt{ {(v_{x})}^{2} + {(b \times v_{x})}^{2} }$

$\implies \: v = v_{x} \: \sqrt{1 + {b}^{2} }$

$\implies \: kt = v_{x} \: \sqrt{1 + {b}^{2} }$

$\implies \: v_{x} = \dfrac{kt}{ \sqrt{1 + {b}^{2} } }$

So, x component of velocity in time t is given as :

$\boxed{ \bf\: v_{x} = \dfrac{kt}{ \sqrt{1 + {b}^{2} } } }$