The acceleration of a particle, given by relation, `a=-5omega^2 sin(omegat)`. At t=0, x=0 and `v= 5omega`. The displacement of this paticle at time t will be

Options

A. `-5 sin (omegat)`
B. `5 sin (omegat)`
C. `5 Cos (omegat)`
D. `-5 Cos (omegat)`

Answer

To solve the problem step by step, we need to find the displacement of the particle at time \( t \) given the acceleration function and initial conditions.

### Given:
- Acceleration: \( a = -5\omega^2 \sin(\omega t) \)
- At \( t = 0 \): \( x = 0 \) and \( v = 5\omega \)

### Step 1: Relate acceleration to velocity
Acceleration is defined as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = -5\omega^2 \sin(\omega t) \]

### Step 2: Separate variables and integrate
We can rearrange the equation to separate variables:
\[ dv = -5\omega^2 \sin(\omega t) dt \]
Now, we integrate both sides. The limits for \( v \) will be from \( 5\omega \) to \( v \) and for \( t \) from \( 0 \) to \( t \):
\[ \int_{5\omega}^{v} dv = -5\omega^2 \int_{0}^{t} \sin(\omega t) dt \]

### Step 3: Perform the integration
The left side integrates to:
\[ v - 5\omega \]
The right side integrates as follows:
\[ -5\omega^2 \left[-\frac{1}{\omega} \cos(\omega t) \right]_{0}^{t} = -5\omega^2 \left[-\frac{1}{\omega} (\cos(\omega t) - \cos(0))\right] \]
This simplifies to:
\[ -5\omega^2 \left[-\frac{1}{\omega} (\cos(\omega t) - 1)\right] = 5\omega(\cos(\omega t) - 1) \]
Thus, we have:
\[ v - 5\omega = 5\omega(\cos(\omega t) - 1) \]
Rearranging gives:
\[ v = 5\omega \cos(\omega t) \]

### Step 4: Relate velocity to displacement
Now, we know that:
\[ v = \frac{ds}{dt} \]
Thus,
\[ \frac{ds}{dt} = 5\omega \cos(\omega t) \]
We separate variables again:
\[ ds = 5\omega \cos(\omega t) dt \]

### Step 5: Integrate to find displacement
Integrate both sides:
\[ \int_{0}^{s} ds = 5\omega \int_{0}^{t} \cos(\omega t) dt \]
The left side integrates to \( s \), and the right side integrates as follows:
\[ 5\omega \left[\frac{1}{\omega} \sin(\omega t)\right]_{0}^{t} = 5\sin(\omega t) \]
Thus, we have:
\[ s = 5\sin(\omega t) \]

### Step 6: Add the linear term from initial conditions
Since the initial displacement at \( t = 0 \) is \( 0 \), we can add the linear term from the velocity:
\[ s = 5\sin(\omega t) + 5\omega t \]

### Final Answer:
The displacement of the particle at time \( t \) is:
\[ s(t) = 5\sin(\omega t) + 5\omega t \]