If `s=e^(t)` (sin t - cos t) is the equation of motion of a moving particle, then acceleration at time t is given by

Options

A. `2 e^(t)` (cos t+sint)
B. `2 e^(t)` (cos t-sint)
C. `e^(t)` (cos t-sint)
D. `e^(t)` (cos t+sint)

Answer

To find the acceleration of the particle given the equation of motion \( s = e^t (\sin t - \cos t) \), 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 \( v \).

Given:
\[ s = e^t (\sin t - \cos t) \]

Using the product rule of differentiation, which states that if \( u \) and \( v \) are functions of \( t \), then:
\[ \frac{d}{dt}(uv) = u'v + uv' \]

Let:
- \( u = e^t \) and \( v = \sin t - \cos t \)

Now, we differentiate:
- \( u' = e^t \)
- \( v' = \cos t + \sin t \) (since the derivative of \( \sin t \) is \( \cos t \) and the derivative of \( -\cos t \) is \( \sin t \))

Applying the product rule:
\[ v = \frac{ds}{dt} = u'v + uv' = e^t (\sin t - \cos t) + e^t (\cos t + \sin t) \]

### Step 2: Simplify the expression for velocity
Now, we can simplify the expression:
\[ v = e^t (\sin t - \cos t) + e^t (\cos t + \sin t) \]
Combining like terms:
\[ v = e^t ((\sin t - \cos t) + (\cos t + \sin t)) = e^t (2\sin t) \]

### Step 3: Differentiate the velocity to find acceleration
Next, we differentiate the velocity \( v \) to find the acceleration \( a \).

\[ a = \frac{dv}{dt} = \frac{d}{dt}(e^t (2\sin t)) \]

Again, we will use the product rule:
Let:
- \( u = e^t \) and \( v = 2\sin t \)

Now, we differentiate:
- \( u' = e^t \)
- \( v' = 2\cos t \)

Applying the product rule:
\[ a = u'v + uv' = e^t (2\sin t) + e^t (2\cos t) \]

### Step 4: Simplify the expression for acceleration
Combining the terms:
\[ a = e^t (2\sin t) + e^t (2\cos t) = 2e^t (\sin t + \cos t) \]

### Final Result
Thus, the acceleration at time \( t \) is given by:
\[ a = 2e^t (\sin t + \cos t) \]

---