The displacement of a particle at time t is given by ` s= 2t^(3) -5t^(2)+ 4t-3` . Find the velocity when t = 2 sec.

Answer

To find the velocity of the particle at time \( t = 2 \) seconds, we start with the given displacement function:

\[ s(t) = 2t^3 - 5t^2 + 4t - 3 \]

### Step 1: Differentiate the displacement function
To find the velocity, we need to differentiate the displacement function \( s(t) \) with respect to time \( t \):

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

### Step 2: Apply the power rule
Using the power rule of differentiation, we differentiate each term:

- The derivative of \( 2t^3 \) is \( 6t^2 \) (since \( 3 \times 2 = 6 \) and we reduce the power by 1).
- The derivative of \( -5t^2 \) is \( -10t \) (since \( 2 \times -5 = -10 \)).
- The derivative of \( 4t \) is \( 4 \).
- The derivative of a constant \( -3 \) is \( 0 \).

So, we have:

\[ v(t) = 6t^2 - 10t + 4 \]

### Step 3: Substitute \( t = 2 \) seconds into the velocity function
Now, we substitute \( t = 2 \) into the velocity function to find the velocity at that moment:

\[ v(2) = 6(2^2) - 10(2) + 4 \]

### Step 4: Calculate the values
Now we calculate each term:

- \( 2^2 = 4 \)
- \( 6 \times 4 = 24 \)
- \( 10 \times 2 = 20 \)

Putting it all together:

\[ v(2) = 24 - 20 + 4 \]

### Step 5: Simplify the expression
Now, simplify the expression:

\[ v(2) = 24 + 4 - 20 = 28 - 20 = 8 \]

### Conclusion
Thus, the velocity of the particle at \( t = 2 \) seconds is:

\[ \boxed{8 \text{ m/s}} \]

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