When `xpropt^(n)`, acceleration is constant when n equals

Options

A. greater than 2
B. less than zero
C. 2
D. nothing can be divided

Answer

To determine the value of \( n \) for which the acceleration is constant when \( x \) is proportional to \( t^n \), we can follow these steps:

### Step 1: Understand the relationship
Given that \( x \) (displacement) is proportional to \( t^n \), we can express this as:
\[ x = k \cdot t^n \]
where \( k \) is a constant.

### Step 2: Find the velocity
Velocity \( v \) is the rate of change of displacement with respect to time, which can be expressed as:
\[ v = \frac{dx}{dt} \]
Differentiating \( x = k \cdot t^n \) with respect to \( t \):
\[ v = \frac{d}{dt}(k \cdot t^n) = k \cdot n \cdot t^{n-1} \]

### Step 3: Find the acceleration
Acceleration \( a \) is the rate of change of velocity with respect to time, which can be expressed as:
\[ a = \frac{dv}{dt} \]
Now, differentiating \( v = k \cdot n \cdot t^{n-1} \) with respect to \( t \):
\[ a = \frac{d}{dt}(k \cdot n \cdot t^{n-1}) = k \cdot n \cdot (n-1) \cdot t^{n-2} \]

### Step 4: Determine the condition for constant acceleration
For acceleration to be constant, it should not depend on \( t \). This means the term \( t^{n-2} \) must be independent of \( t \). The only way for this to happen is if the exponent \( n-2 = 0 \), which leads to:
\[ n - 2 = 0 \implies n = 2 \]

### Conclusion
Thus, the value of \( n \) for which the acceleration is constant is:
\[ \boxed{2} \]