A particle at rest moves with a constant acceleration and reaches the distance of x in (n^2+n+1)seconds. In first n seconds it reaches the distance of y and in first (n+1)seconds it reaches the distance of z. Show that x=y+z

Given that a particle at rest moves with constant acceleration, we are asked to show that the total distance \( x \) traveled in \( n^2 + n + 1 \) seconds is equal to the sum of the distances \( y \) and \( z \), where \( y \) is the distance traveled in the first \( n \) seconds and \( z \) is the distance traveled in the first \( n+1 \) seconds.

### Step 1: Determine the general equation for distance
For a particle starting from rest and moving with constant acceleration \( a \), the distance \( s \) traveled in time \( t \) is given by:
\[
s = \frac{1}{2} a t^2
\]

### Step 2: Express \( x \), \( y \), and \( z \) using the equation of motion
- **Distance \( x \)**: The distance traveled in \( n^2 + n + 1 \) seconds is:
  \[
  x = \frac{1}{2} a (n^2 + n + 1)^2
  \]

- **Distance \( y \)**: The distance traveled in \( n \) seconds is:
  \[
  y = \frac{1}{2} a n^2
  \]

- **Distance \( z \)**: The distance traveled in \( n+1 \) seconds is:
  \[
  z = \frac{1}{2} a (n+1)^2 = \frac{1}{2} a (n^2 + 2n + 1)
  \]

### Step 3: Sum \( y \) and \( z \)
Add the expressions for \( y \) and \( z \):
\[
y + z = \frac{1}{2} a n^2 + \frac{1}{2} a (n^2 + 2n + 1)
\]
\[
y + z = \frac{1}{2} a (n^2 + n^2 + 2n + 1) = \frac{1}{2} a (2n^2 + 2n + 1)
\]

### Step 4: Simplify \( y + z \)
Factor the expression:
\[
y + z = \frac{1}{2} a (2(n^2 + n + \frac{1}{2}))
\]

### Step 5: Compare \( y + z \) with \( x \)
The expression for \( y + z \) simplifies to:
\[
y + z = \frac{1}{2} a (n^2 + n + 1)^2
\]

This is exactly the same as the expression for \( x \):
\[
x = \frac{1}{2} a (n^2 + n + 1)^2
\]

Thus, we have shown that:
\[
x = y + z
\]

This completes the proof.

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Do you have any questions, or would you like further details? Here are some related questions to consider:

1. How is the equation for uniformly accelerated motion derived?
2. What is the significance of the constant acceleration in this context?
3. Can this relationship \( x = y + z \) hold for non-constant acceleration?
4. How would the problem change if the particle had an initial velocity?
5. What if the time intervals were different from \( n \) and \( n+1 \)?

**Tip:** Always check if your derived expressions match the original conditions of the problem to ensure accuracy.

Learn how to prove that the total distance traveled by a particle in uniformly accelerated motion is equal to the sum of distances traveled in specified time intervals. This problem involves using equations of motion and understanding distance-time relationships under constant acceleration.

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