Is the answer 50^(1/2) differentiate x(t) dx/dt=v(x component of velocity)=50cos10t dy/dt=v(y component of velocity)=-50sin10t and take the magnitude ((50cos10t)^2 +(50sin10t)^2)^(1/2) you will get the speed of the particle

Point O is a?
In order to determine what point O represents, we need to analyze the given position function x = x(t).

Position Function: x = 5sin(10t)

From the given position function, we can see that the object's position is dependent on time. The object's position along the x-axis is given by the sine function with an amplitude of 5 and a period of 2π/10 or π/5.

Explanation:
To understand the nature of point O, we need to analyze the given position function and its graphical representation.

1. Amplitude:
The amplitude of the sine function is the maximum distance from the mean position. In this case, the amplitude is 5, indicating that the object oscillates between -5 and 5 along the x-axis.

2. Period:
The period of the sine function is the time taken to complete one full cycle. In this case, the period is π/5, indicating that the object completes one full oscillation along the x-axis in π/5 units of time.

3. Phase Shift:
The phase shift of the sine function determines the displacement of the graph along the x-axis. In this case, there is no phase shift mentioned in the given function, so the graph starts at the mean position (x = 0) at t = 0.

4. Graphical Representation:
Plotting the given position function on a graph with time (t) on the x-axis and position (x) on the y-axis, we can observe the following characteristics:

- The graph is a sinusoidal wave centered around the x-axis.
- The graph oscillates between -5 and 5 along the x-axis.
- The graph completes one full cycle in π/5 units of time.

Conclusion:
Based on the analysis of the given position function x = 5sin(10t), we can conclude that point O represents the mean position of the object, which is at x = 0. This means that when the object is at point O, it is at the center of its oscillation along the x-axis.