The distance travelled by an object along a straight line in time t is given by `s = 3 -41 + 5t^(2)` , the initial velocity of the objectis

Options

A. 3 unit
B. `-3` unit
C. 4 unit
D. `-4` units

Explanation

To find the initial velocity of the object given the distance traveled as a function of time \( s(t) = 3 - 4t + 5t^2 \), we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Equation**: The distance \( s \) is given as a function of time \( t \):
\[ s(t) = 3 - 4t + 5t^2 \]

2. **Differentiate the Distance Function**: To find the velocity, we need to differentiate the distance function with respect to time \( t \). The velocity \( v(t) \) is given by:
\[ v(t) = \frac{ds}{dt} \]

3. **Calculate the Derivative**:
- The derivative of \( 3 \) is \( 0 \).
- The derivative of \( -4t \) is \( -4 \).
- The derivative of \( 5t^2 \) is \( 10t \).

Therefore, the velocity function is:
\[ v(t) = 0 - 4 + 10t = -4 + 10t \]

4. **Find the Initial Velocity**: The initial velocity is the velocity at \( t = 0 \):
\[ v(0) = -4 + 10(0) = -4 \]

5. **State the Result**: The initial velocity of the object is:
\[ v(0) = -4 \text{ units} \]

### Final Answer:
The initial velocity of the object is \( -4 \) units.

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