Example of x0 in the kinematic equation for displacement

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e-zero
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In the kinematic equation x = x0 + v0 * t + .5 * a * t^2, can someone give me an example in which x0 has a value that is not zero?
 
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Hi e-zero! ##x_{0}## is just the initial position of the test particle that you are looking at the trajectory of under constant acceleration; similarly ##v_{0}## is just the initial velocity of the trajectory. You can choose them to be whatever you want them to be. I can for example drop my object from a height of 5 meters from the ground, starting at rest, in which case ##x_{0} = 5 \text{m}## , ##v_{0} = 0 \text{m /s}## , and (ignoring air resistance) the equations of motion become ##x(t) = 5 \text{m} - \frac{1}{2}gt^{2}##. Hope that helps friend!
 
Ok, is there an example where both x and x0 are both not zero?
 
##x(t)## is a function of time so it will change its value at every instant of time. In the example I gave above, ##x(t)## is the height of the particle from the ground so it will be non-zero up until the first time the particle hits the ground.
 
Can you give an example that does not involve vertical, but horizontal motion in which x and x0 are both not zero?
 
Sure! Choose an origin and take a particle located 5 meters horizontally from that origin and give it an initial kick of 5 meters per second in the horizontal direction.
 
Ok. I was just confusing myself. It all depends on the question if you should take x0 to be zero or not.
 
e-zero said:
Ok. I was just confusing myself. It all depends on the question if you should take x0 to be zero or not.

Not really. It depends on where you want the origin of your system of coordinates to be. In WBN's example, he chose the ground as origin. That's why x_0 in his case was 5m. For the same example he could have chosen its origin 5 meters above the ground, in which case x_0 would have been 0.