Instantaneous Velocity and Acceleration

Let \(\vec{v}\) and \(\vec{a}\) be instantaneous velocity and the acceleration respectively of a particle moving in a plane. The rate of change of speed (dv/dt) of the particle is:
(a) \(|a|\)
(b) \((v.a)/|v|\)
(c) the component of \(\vec{a}\) in the direction of \(\vec{v}\)
(d) the component of \(\vec{a}\) perpendicular to \(\vec{v}\)

Solution:

Let us consider \(v^2=v_x^2+v_y^2\).
We differentiate the above equation.
\(\frac{dv}{dt}\)=\((v_xa_x+v_ya_y)v\)=\(\frac{v.a}{v}\).
Hence, the correct option will be B along with C since the component of a is in the direction of v.

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