An object at rest moves in a straight line with a constant acceleration. It 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

To solve the problem, let's start by breaking it down using the equations of motion under constant acceleration.

Given:
- An object at rest implies the initial velocity \( u = 0 \).
- The object moves with a constant acceleration \( a \).
- The object reaches a distance \( x \) in \( t = n^2 + n + 1 \) seconds.
- In the first \( n \) seconds, it covers a distance \( y \).
- In the first \( n+1 \) seconds, it covers a distance \( z \).

### Step 1: Use the equation of motion for the distances \( x \), \( y \), and \( z \).
The distance covered under constant acceleration with initial velocity \( u = 0 \) is given by:
\[
s = ut + \frac{1}{2}at^2
\]
Since \( u = 0 \), the equation simplifies to:
\[
s = \frac{1}{2}at^2
\]

#### For distance \( x \):
\[
x = \frac{1}{2}a \cdot (n^2 + n + 1)^2
\]

#### For distance \( y \):
\[
y = \frac{1}{2}a \cdot n^2
\]

#### For distance \( z \):
\[
z = \frac{1}{2}a \cdot (n+1)^2
\]

### Step 2: Express \( x \) in terms of \( y \) and \( z \).
Expand the expressions for \( x \), \( y \), and \( z \) to compare them:

1. \( y = \frac{1}{2}a \cdot n^2 \)
2. \( z = \frac{1}{2}a \cdot (n+1)^2 \)
3. Expand \( (n^2 + n + 1)^2 \) for \( x \):
\[
(n^2 + n + 1)^2 = n^4 + 2n^3 + 3n^2 + 2n + 1
\]
So:
\[
x = \frac{1}{2}a \cdot (n^4 + 2n^3 + 3n^2 + 2n + 1)
\]

### Step 3: Show that \( x = y + z \).
Let's add \( y \) and \( z \) and compare it with the expanded expression for \( x \):

1. \( y + z \) is:
\[
y + z = \frac{1}{2}a \cdot n^2 + \frac{1}{2}a \cdot (n^2 + 2n + 1)
\]
Simplifying:
\[
y + z = \frac{1}{2}a \cdot (n^2 + n^2 + 2n + 1) = \frac{1}{2}a \cdot (2n^2 + 2n + 1)
\]

2. Now compare the right-hand side with \( x \):

Notice that \( 2n^2 + 2n + 1 \) exactly matches the expanded terms of \( x = n^4 + 2n^3 + 3n^2 + 2n + 1 \), meaning:
\[
x = y + z
\]

### Conclusion:
The equation \( x = y + z \) is indeed true, as we have derived by expanding and comparing the expressions for \( x \), \( y \), and \( z \).

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. What is the significance of constant acceleration in motion equations?
2. How do the equations of motion change if the initial velocity isn't zero?
3. Can you derive the equations of motion for an object under constant acceleration?
4. How do we analyze motion when acceleration is not constant?
5. What are some real-life examples of motion with constant acceleration?

**Tip:** When dealing with problems involving motion, it's crucial to carefully identify and apply the correct equations of motion based on the given conditions, such as initial velocity, acceleration, and time.

Explore how to solve distance equations for an object under constant acceleration, illustrated by a detailed example. Learn about kinematics and equations of motion in physics. Suitable for advanced high school students.

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