If `S=t^(4)-5t^(2)+8t-3`, then initial velocity of the particle is

Options

A. 0
B. 8 units / sec
C. 6 units/sec
D. 5 units / sec

Answer

To find the initial velocity of the particle given the function \( S(t) = t^4 - 5t^2 + 8t - 3 \), we will follow these steps:

### Step 1: Find the velocity function
The velocity \( v(t) \) is the derivative of the displacement function \( S(t) \) with respect to time \( t \). We need to differentiate \( S(t) \).

\[ v(t) = \frac{dS}{dt} = \frac{d}{dt}(t^4 - 5t^2 + 8t - 3) \]

### Step 2: Differentiate the function
Now, we differentiate each term in the function:

- The derivative of \( t^4 \) is \( 4t^3 \).
- The derivative of \( -5t^2 \) is \( -10t \).
- The derivative of \( 8t \) is \( 8 \).
- The derivative of the constant \( -3 \) is \( 0 \).

Putting it all together, we get:

\[ v(t) = 4t^3 - 10t + 8 \]

### Step 3: Calculate the initial velocity
The initial velocity is the velocity at \( t = 0 \). We substitute \( t = 0 \) into the velocity function:

\[ v(0) = 4(0)^3 - 10(0) + 8 = 0 - 0 + 8 = 8 \]

### Final Answer
The initial velocity of the particle is \( 8 \) units per second.

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