Question

1.42. A particle moves along the plane trajectory y (x) with velocity
v whose modulus is constant. Find the acceleration of the particle
at the point x = 0 and the curvature radius of the trajectory
at that point if the trajectory has the form
(a) of a parabola y = ax2;

Answer 1

Given:-

⠀⠀⠀

• Plane trajectory = y(x)
• Velocity = v
• Modulus = Constant

⠀⠀⠀

To Find:-

⠀⠀⠀

• Acceleration of particle at point where x = 0
• Curvature radius of trajectory if trajectory has the form of a probably y = ax²

⠀⠀⠀

Solution:-

⠀⠀⠀

➥ As given in question,

y = ax²

⠀⠀⠀

➥ Let us differentiate twice the path of equation [ y(x)] with respect to time

$\tt{\frac{dy}{dt} = 2ax\frac{dx}{dt}}$

⠀⠀⠀

Now, take $\tt{\frac{dy}{dt}}$ as vᵧ and $\tt{\frac{dx}{dt}}$ as vₓ

⠀⠀⠀

So,

➭ vᵧ = 2axvₓ

➭ aᵧ = 2axaₓ + 2avₓ²

⠀⠀⠀

➥ As given in question x = 0

So, vᵧ = 0 ; vₓ = v

Hence,

aᵧ = 2av²

⠀⠀⠀

➥ As given in question that is speed is constant sothe tangentical acceleration is zero.

Hence,

aₓ = $\tt{\frac{dv}{dt}}$

➭ Net acceleration = 2av²

⠀⠀⠀

⠀⠀⠀

➥ Radius of curvature

= $\tt{\frac{velocity²}{net\ acceleration}}$

= $\tt{\frac{v²}{2av²}}$

= $\tt{\frac{\not{v²}}{2a\not{v²}}}$

= $\tt{\frac{1}{2a}}$

⠀⠀⠀

Answer:-

⠀⠀⠀

• Acceleration of particle = 2av²

• Radius of curvature = $\tt{\frac{1}{2a}}$

════◄••❀••►════

Answer 2

Answer:

Given:-

⠀⠀⠀

Plane trajectory = y(x)

Velocity = v

Modulus = Constant

⠀⠀⠀

To Find:-

⠀⠀⠀

Acceleration of particle at point where x = 0

Curvature radius of trajectory if trajectory has the form of a probably y = ax²

⠀⠀⠀

Solution:-

⠀⠀⠀

➥ As given in question,

y = ax²

⠀⠀⠀

➥ Let us differentiate twice the path of equation [ y(x)] with respect to time

\tt{\frac{dy}{dt} = 2ax\frac{dx}{dt}}

dt

dy

=2ax

dt

dx

⠀⠀⠀

Now, take \tt{\frac{dy}{dt}}

dt

dy

as vᵧ and \tt{\frac{dx}{dt}}

dt

dx

as vₓ

⠀⠀⠀

So,

➭ vᵧ = 2axvₓ

➭ aᵧ = 2axaₓ + 2avₓ²

⠀⠀⠀

➥ As given in question x = 0

So, vᵧ = 0 ; vₓ = v

Hence,

aᵧ = 2av²

⠀⠀⠀

➥ As given in question that is speed is constant sothe tangentical acceleration is zero.

Hence,

aₓ = \tt{\frac{dv}{dt}}

dt

dv

➭ Net acceleration = 2av²

⠀⠀⠀

⠀⠀⠀

➥ Radius of curvature

= \tt{\frac{velocity²}{net\ acceleration}}

net acceleration

velocity²

= \tt{\frac{v²}{2av²}}

2av²

v²

= \tt{\frac{\not{v²}}{2a\not{v²}}}

2a



v²



v²

= \tt{\frac{1}{2a}}

2a

1

⠀⠀⠀

Answer:-

⠀⠀⠀

Acceleration of particle = 2av²

Radius of curvature = \tt{\frac{1}{2a}}

2a

1

════◄••❀••►════

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