The displacement `s` of a point moving in a straight line is given by:
`s = 8 t^(2) + 3t - 5`
`s` being in cm and `t` in s. The initial velocity of the particle is:

Options

A. 3 cm/s
B. 16 cm/s
C. 19 cm/s
D. zero

Explanation

To find the initial velocity of the particle given the displacement equation \( s = 8t^2 + 3t - 5 \), we will follow these steps:

### Step 1: Understand the relationship between displacement and velocity
The velocity \( v \) of a particle is the derivative of displacement \( s \) with respect to time \( t \). This means we need to differentiate the displacement function.

### Step 2: Differentiate the displacement function
Given the displacement function:
\[ s = 8t^2 + 3t - 5 \]
we differentiate \( s \) with respect to \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(8t^2 + 3t - 5) \]

### Step 3: Apply the differentiation rules
Using the power rule of differentiation:
- The derivative of \( 8t^2 \) is \( 16t \).
- The derivative of \( 3t \) is \( 3 \).
- The derivative of a constant (-5) is \( 0 \).

Thus, we have:
\[ v = 16t + 3 \]

### Step 4: Calculate the initial velocity
The initial velocity is the velocity at time \( t = 0 \):
\[ v(0) = 16(0) + 3 = 3 \, \text{cm/s} \]

### Conclusion
The initial velocity of the particle is:
\[ \boxed{3 \, \text{cm/s}} \]

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