A particle moves along the parabola y = x^2 such
that the x-component of velocity is always
2m/s. The acceleration of the particle is
To find the acceleration of the particle, we need to differentiate the given equation of motion twice with respect to time.
Given: y = x^2
Differentiating with respect to time (t), we get:
dy/dt = d(x^2)/dt
Here, dy/dt represents the derivative of y with respect to t, and d(x^2)/dt represents the derivative of x^2 with respect to t.
Since the x-component of velocity is always 2 m/s, we can write:
dx/dt = 2
Now, differentiating both sides of the equation with respect to t again, we get:
d^2y/dt^2 = d(dx/dt)/dt
d^2y/dt^2 represents the second derivative of y with respect to t, and d(dx/dt)/dt represents the derivative of dx/dt with respect to t.
Using the chain rule of differentiation, we can simplify the equation further:
d^2y/dt^2 = d(2)/dt
Differentiating a constant (2) with respect to t gives us 0.
Therefore, the acceleration of the particle is 0.