A particle moving in a straight line with constant acceleration passing over a distance x,y,z in equal consecutive intervals of time (t). find the relations between x,y,z. Find the acceleration of the particle.

To find the relations between x, y, and z, we can use the equations of motion. Let's assume the initial velocity of the particle is u, its final velocity is v, and the acceleration is a. Since the particle is moving with constant acceleration, we can use the following equations:

1. v = u + at ---(1)
2. s = ut + (1/2)at^2 ---(2)

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time interval
s = distance traveled

In this case, the particle passes over the distances x, y, and z in equal consecutive intervals of time (t). So, for each distance, the time interval is the same, i.e., t.

Using equation (2) for x, y, and z, we have:

For distance x:
x = ut + (1/2)at^2

For distance y:
y = (u + at)t = ut + at^2

For distance z:
z = (u + 2at)t = ut + 2at^2

Now, let's find the relations between x, y, and z:

1. Multiply the equation for x by 2 and subtract the equation for z from it:
2x - z = 2(ut + (1/2)at^2) - (ut + 2at^2)
2x - z = ut

2. Rearrange the equation for y:
y = ut + at^2

From our calculation above, we see that 2x - z = ut. This implies that y = ut + at^2 = 2x - z.

Therefore, the relation between x, y, and z is y = 2x - z.

To find the acceleration of the particle, we can rearrange equation (1) as follows:

v = u + at

Since the particle starts from rest (u = 0), the final velocity v is the velocity acquired after time t. Therefore:

v = at

Hence, the acceleration of the particle is a = v/t.