To compute the acceleration of the trolley, we need to first find its velocity vector, and then take the derivative of the velocity vector with respect to time to find the acceleration vector.
At point O, the trolley is at the bottom of the parabolic track, where y = 0. Therefore, the position vector of the trolley is given by r = xi, where i is the unit vector in the x-direction. The velocity vector of the trolley is given by the derivative of the position vector with respect to time:
v = dr/dt = (d/dt)(xi) = (dx/dt)i
Since the trolley is traveling at a constant speed of 40 mi/h, its velocity vector has a magnitude of 40 mi/h. To find the x-component of the velocity vector, we can use the fact that the speed is constant:
40 mi/h = |v| = sqrt[(dx/dt)^2]
Solving for dx/dt, we get:
dx/dt = 40/sqrt(mi^2/h^2) = 40/5280 ft/s
Therefore, the velocity vector of the trolley at point O is given by:
v = (40/5280)i ft/s
Taking the derivative of the velocity vector with respect to time, we get the acceleration vector:
a = dv/dt = (d/dt)(40/5280)i = 0
Therefore, the acceleration of the trolley at point O is zero.
At point A, the position vector of the trolley is given by:
r = xi + (x^2/500)j
where j is the unit vector in the y-direction. To find the velocity vector of the trolley, we need to take the derivative of the position vector with respect to time:
v = dr/dt = (d/dt)(xi) + (d/dt)((x^2/500)j)
The x-component of the velocity vector is given by:
dx/dt = 40 mi/h = 40/5280 ft/s
The y-component of the velocity vector is given by:
(dy/dt)j = (d/dt)((x^2/500)j) = (2x/500)(dx/dt)j
At point A, x = 200 ft (since y = x^2/500 = 40000/500 = 80 ft at point A), so we can compute the y-component of the velocity vector:
(dy/dt)j = (2x/500)(dx/dt)j = (2*200/500)(40/5280)j = (16/1320)j ft/s
Therefore, the velocity vector of the trolley at point A is given by:
v = (40/5280)i + (16/1320)j ft/s
Taking the derivative of the velocity vector with respect to time, we get the acceleration vector:
a = dv/dt = (d/dt)((40/5280)i) + (d/dt)((16/1320)j)
The x-component of the acceleration vector is zero, since the speed of the trolley is constant:
(d/dt)((40/5280)i) = 0
The y-component of the acceleration vector is given by:
(d/dt)((16/1320)j) = (d/dt)((2x/500)(dx/dt)j) = [(2/500)(d^2x/dt^2) + (2/500)(dx/dt)^2]j
To find the second derivative of x with respect to time, we can use the fact that the speed is constant: