If `s=ae^(t) + be^(-t)` is the equation of motion of a particle, then its acceleration is equal to

Options

A. s
B. 2s
C. 3s
D. 4s

Answer

To find the acceleration of the particle given the equation of motion \( s = ae^t + be^{-t} \), we will follow these steps:

### Step 1: Differentiate the displacement to find velocity
The first step is to differentiate the displacement \( s \) with respect to time \( t \) to find the velocity \( v \).

\[ v = \frac{ds}{dt} = \frac{d}{dt}(ae^t + be^{-t}) \]

Using the rules of differentiation, we get:

\[ v = a \frac{d}{dt}(e^t) + b \frac{d}{dt}(e^{-t}) = ae^t - be^{-t} \]

### Step 2: Differentiate the velocity to find acceleration
Next, we differentiate the velocity \( v \) with respect to time \( t \) to find the acceleration \( a \).

\[ a = \frac{dv}{dt} = \frac{d}{dt}(ae^t - be^{-t}) \]

Again, applying the rules of differentiation:

\[ a = a \frac{d}{dt}(e^t) - b \frac{d}{dt}(e^{-t}) = ae^t + be^{-t} \]

### Step 3: Relate acceleration to displacement
Notice that the expression we obtained for acceleration \( a = ae^t + be^{-t} \) is the same as the original displacement \( s \).

Thus, we can conclude:

\[ a = s \]

### Final Answer
The acceleration of the particle is equal to the displacement \( s \).

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