Always be accelerated

Classical Mechanics Level 2

A particle moves along the parabolic path y = a x 2 y = ax^2 in such a way that the x x component of velocity remains constant, say c c . What is the acceleration of the particle?

3 c 3c a c 2 ac^2 a 2 c a^2c 2 a c 2 2ac^2

Aayushyamaan Raj by AAYUSHYAMAAN RAJ , 21, India

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1 solution

Tom Engelsman
Apr 29, 2018

If the x-component of the particle's velocity is constant, then its x-position is equal to x = c t x = ct and its y-position is y = a c 2 t 2 y = ac^2t^2 . If we are interested in the particle's acceleration (whose components are equal to a x ( t ) = x ′ ′ ( t ) ; a y ( t ) = y ′ ′ ( t ) a_{x}(t) = x''(t); a_{y}(t) = y''(t) , then we have:

x ′ ′ ( t ) = 0 ; y ′ ′ ( t ) = 2 a c 2 x''(t) = 0; y''(t) = 2ac^2

and the magnitude of the acceleration calculates to ∣ a ∣ = a x ( t ) 2 + a y ( t ) 2 = 0 2 + ( 2 a c 2 ) 2 = 2 a c 2 . |a| = \sqrt{a_{x}(t)^2 + a_{y}(t)^2} = \sqrt{0^2 + (2ac^2)^2} = \boxed{2ac^2}.

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