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A particle is moving in a parabolic path y=4x^2 with constant speed v m/s. The acceleration of particle when it crosses the origin is?

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User Jonh Doe
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Final answer:

The acceleration of the particle when it crosses the origin is 0 m/s^2.

Step-by-step explanation:

To find the acceleration of a particle moving in a parabolic path, we need to differentiate the equation of motion twice with respect to time. The equation of motion for the particle is given by y = 4x^2, where y is the position in the vertical direction and x is the position in the horizontal direction. Taking the derivative of this equation twice gives us the acceleration of the particle.

Differentiating the equation y = 4x^2 with respect to time, we get:

v = d(4x^2)/dt

Using the chain rule, we have:

dv/dt = d(4x^2)/dx * dx/dt

Since the particle is moving with constant speed v, dx/dt = v. Therefore, the acceleration of the particle when it crosses the origin is:

dv/dt = d(4x^2)/dx * v

Now, taking the derivative of 4x^2 with respect to x, we get:

dv/dt = 8x * v

Substituting x = 0 (since the particle crosses the origin), we have:

dv/dt = 8 * 0 * v = 0

Therefore, the acceleration of the particle when it crosses the origin is 0 m/s^2.

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User XDaevax
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