A particle moves with constant speed of 3m/sec along path y=3x^2
What is the acceleration of the particle at x=1.5m?
Since the speed is constant, the acceleration must be caused by the change in direction. The angle of the trajectory to the x axis is
theta = arctan (dy/dx) = arctan (6x)
When x = 1.5, theta = arctan 9 = 83.66 degrees
The acceleration will be
(speed)x d(theta)/dt
= (speed)* d(theta)/dx* dx/dt
Take it from there.
To find the acceleration of the particle at a given point on the path, we need to differentiate the path equation with respect to time (assuming x is a function of time).
Given the path equation y = 3x^2, we can differentiate this equation with respect to time using the chain rule.
dy/dt = d(3x^2)/dt
To differentiate x^2 with respect to time, we need to apply the chain rule, as x is a function of time.
dx/dt * d(x^2)/dx
Since the particle moves with constant speed, dx/dt is a constant value, let's call it v, which is equal to 3 m/s.
So, dy/dt = v * d(x^2)/dx
Now, we can differentiate x^2 with respect to x.
dy/dt = v * 2x
At x = 1.5 m,
dy/dt = v * 2 * (1.5)
dy/dt = 2v * 1.5
dy/dt = 3v
So, the velocity of the particle at x = 1.5 m is 3v m/s.
Since acceleration is the rate of change of velocity, we differentiate velocity with respect to time.
a = d(3v)/dt
Since v is constant, its derivative with respect to time is zero.
a = 0
Therefore, the acceleration of the particle at x = 1.5 m is 0 m/s^2.
To find the acceleration of the particle at a given position, we need to determine the second derivative of the position function with respect to time (a = d^2y/dt^2). However, in this case, we are given the position function as a function of x instead of time.
To find the acceleration at x = 1.5m, we first need to express the position function in terms of time. Given that the particle moves with a constant speed of 3 m/s, we can relate x and t using the equation s = v*t, where s represents the displacement.
Since the particle moves along the path y = 3x^2, we can differentiate this equation with respect to time to get dy/dt in terms of dx/dt. Differentiating the equation y = 3x^2 with respect to time, we get:
dy/dt = d(3x^2)/dt = 6x * dx/dt
Since we know the particle moves with a constant speed, dx/dt is equal to the constant speed, which is 3 m/s. Therefore, we have:
dy/dt = 6x * 3
Now, we can find dy/dt in terms of time by replacing x with the value at which we want to find the acceleration, which is x = 1.5m:
dy/dt = 6 * 1.5 * 3 = 27 m/s
Now, we need to find the acceleration, which is given by the second derivative of y with respect to t. Since y = 3x^2, we can differentiate this equation with respect to time twice. Differentiating dy/dt = 27 with respect to time, we get:
d^2y/dt^2 = d(27)/dt = 0
Therefore, the acceleration of the particle at x = 1.5m is 0 m/s^2.