The equation of motion of a particle started at t=0 is given by `x=5sin(20t+pi/3)`, where x is in centimetre and t in second. When does the particle
a. first come rest
b. first have zero acceleration
c. first have maximum speed?

Explanation

To solve the problem step by step, we will analyze the motion of the particle described by the equation \( x = 5 \sin(20t + \frac{\pi}{3}) \).

### Step 1: Identify Parameters
From the equation, we can identify:
- Amplitude \( A = 5 \) cm
- Angular frequency \( \omega = 20 \) rad/s
- Initial phase \( \phi = \frac{\pi}{3} \)

### Step 2: Finding When the Particle First Comes to Rest
The particle comes to rest when its velocity is zero. The velocity \( v \) in simple harmonic motion is given by:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
Setting \( v = 0 \):
\[ A \omega \cos(20t + \frac{\pi}{3}) = 0 \]
This implies:
\[ \cos(20t + \frac{\pi}{3}) = 0 \]
The cosine function is zero at:
\[ 20t + \frac{\pi}{3} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
Taking the first instance (n=0):
\[ 20t + \frac{\pi}{3} = \frac{\pi}{2} \]
Solving for \( t \):
\[ 20t = \frac{\pi}{2} - \frac{\pi}{3} \]
Finding a common denominator (6):
\[ 20t = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6} \]
\[ t = \frac{\pi}{120} \text{ seconds} \]

### Step 3: Finding When the Particle First Has Zero Acceleration
The acceleration \( a \) in simple harmonic motion is given by:
\[ a = -A \omega^2 \sin(\omega t + \phi) \]
Setting \( a = 0 \):
\[ -A \omega^2 \sin(20t + \frac{\pi}{3}) = 0 \]
This implies:
\[ \sin(20t + \frac{\pi}{3}) = 0 \]
The sine function is zero at:
\[ 20t + \frac{\pi}{3} = n\pi \quad (n \in \mathbb{Z}) \]
Taking the first instance (n=0):
\[ 20t + \frac{\pi}{3} = 0 \]
This gives:
\[ 20t = -\frac{\pi}{3} \quad \text{(not valid for positive time)} \]
Taking \( n = 1 \):
\[ 20t + \frac{\pi}{3} = \pi \]
Solving for \( t \):
\[ 20t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
\[ t = \frac{2\pi}{60} = \frac{\pi}{30} \text{ seconds} \]

### Step 4: Finding When the Particle First Has Maximum Speed
The maximum speed occurs when the displacement \( x = 0 \):
\[ 0 = 5 \sin(20t + \frac{\pi}{3}) \]
This implies:
\[ \sin(20t + \frac{\pi}{3}) = 0 \]
Using the same reasoning as before:
\[ 20t + \frac{\pi}{3} = n\pi \]
Taking the first instance (n=0):
\[ 20t + \frac{\pi}{3} = 0 \quad \text{(not valid for positive time)} \]
Taking \( n = 1 \):
\[ 20t + \frac{\pi}{3} = \pi \]
Solving for \( t \):
\[ 20t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
\[ t = \frac{2\pi}{60} = \frac{\pi}{30} \text{ seconds} \]

### Summary of Results
- a. The particle first comes to rest at \( t = \frac{\pi}{120} \) seconds.
- b. The particle first has zero acceleration at \( t = \frac{\pi}{30} \) seconds.
- c. The particle first has maximum speed at \( t = \frac{\pi}{30} \) seconds.