The distance in seconds, described by a particle in t seconds is given by `s=ae^t+b/e^t`. The acceleration of the particle at time t is

Options

A. proportional to t
B. Proportional to v
C. s
D. constant

Answer

To find the acceleration of the particle at time \( t \), we start with the given displacement function:

\[ s = ae^t + \frac{b}{e^t} \]

### Step 1: Differentiate the displacement function to find velocity

The first derivative of displacement \( s \) with respect to time \( t \) gives us the velocity \( v \):

\[ v = \frac{ds}{dt} = \frac{d}{dt} \left( ae^t + \frac{b}{e^t} \right) \]

Using the chain rule and the product rule, we differentiate each term:

1. The derivative of \( ae^t \) is \( ae^t \).
2. The derivative of \( \frac{b}{e^t} \) can be rewritten as \( b e^{-t} \). Its derivative is \( -b e^{-t} \).

Thus, we have:

\[ v = ae^t - b e^{-t} \]

### Step 2: Differentiate the velocity function to find acceleration

Now, we differentiate the velocity function \( v \) to find the acceleration \( a \):

\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( ae^t - b e^{-t} \right) \]

Again, using the chain rule:

1. The derivative of \( ae^t \) is \( ae^t \).
2. The derivative of \( -b e^{-t} \) is \( b e^{-t} \) (note the negative sign becomes positive).

Thus, we have:

\[ a = ae^t + b e^{-t} \]

### Final Result

The acceleration of the particle at time \( t \) is:

\[ a = ae^t + b e^{-t} \]