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 described by the equation \( x = 5 \sin(20t + \frac{\pi}{3}) \).

### Step 1: Finding when the particle first comes to rest

1. **Determine the velocity**: The velocity \( v \) is the derivative of the position \( x \) with respect to time \( t \).
\[ v = \frac{dx}{dt} = 5 \cdot 20 \cos(20t + \frac{\pi}{3}) = 100 \cos(20t + \frac{\pi}{3}) \]

2. **Set the velocity to zero**: The particle comes to rest when \( v = 0 \).
\[ 100 \cos(20t + \frac{\pi}{3}) = 0 \]
This implies:
\[ \cos(20t + \frac{\pi}{3}) = 0 \]

3. **Solve for \( t \)**: The cosine function is zero at \( \frac{\pi}{2} + n\pi \) for \( n = 0, 1, 2, \ldots \).
\[ 20t + \frac{\pi}{3} = \frac{\pi}{2} \]
Rearranging gives:
\[ 20t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \]
Thus:
\[ t = \frac{\pi}{120} \text{ seconds} \]

### Step 2: Finding when the particle first has zero acceleration

1. **Determine the acceleration**: The acceleration \( a \) is the derivative of the velocity \( v \).
\[ a = \frac{dv}{dt} = -100 \cdot 20 \sin(20t + \frac{\pi}{3}) = -2000 \sin(20t + \frac{\pi}{3}) \]

2. **Set the acceleration to zero**: The particle has zero acceleration when \( a = 0 \).
\[ -2000 \sin(20t + \frac{\pi}{3}) = 0 \]
This implies:
\[ \sin(20t + \frac{\pi}{3}) = 0 \]

3. **Solve for \( t \)**: The sine function is zero at \( n\pi \) for \( n = 0, 1, 2, \ldots \).
\[ 20t + \frac{\pi}{3} = n\pi \]
For the first instance (n=1):
\[ 20t + \frac{\pi}{3} = \pi \]
Rearranging gives:
\[ 20t = \pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3} \]
Thus:
\[ t = \frac{2\pi}{60} = \frac{\pi}{30} \text{ seconds} \]

### Step 3: Finding when the particle first has maximum speed

1. **Maximum speed occurs at the mean position**: The particle reaches maximum speed when it passes through the mean position (where \( x = 0 \)).

2. **Set the position to zero**:
\[ 5 \sin(20t + \frac{\pi}{3}) = 0 \]
This implies:
\[ \sin(20t + \frac{\pi}{3}) = 0 \]

3. **Solve for \( t \)**: As previously derived, the first instance occurs when:
\[ 20t + \frac{\pi}{3} = \pi \]
Rearranging gives:
\[ t = \frac{\pi}{30} \text{ seconds} \]

### 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.