The displacement s of an object is given as a function of time t by the following equation `s=2t+5t^(2)+3t^(3)`. Calculate the instantaneous velocity of the object at t = 1 s.

Explanation

To find the instantaneous velocity of the object at \( t = 1 \) second given the displacement function \( s(t) = 2t + 5t^2 + 3t^3 \), we will follow these steps:

### Step 1: Differentiate the displacement function
The instantaneous velocity \( v(t) \) is defined as the derivative of the displacement \( s(t) \) with respect to time \( t \). We will differentiate the given function:

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

Differentiating term by term:

- The derivative of \( 2t \) is \( 2 \).
- The derivative of \( 5t^2 \) is \( 10t \).
- The derivative of \( 3t^3 \) is \( 9t^2 \).

Thus, the derivative of the displacement function is:

\[ v(t) = \frac{ds}{dt} = 2 + 10t + 9t^2 \]

### Step 2: Substitute \( t = 1 \) into the velocity function
Now, we will substitute \( t = 1 \) second into the velocity function to find the instantaneous velocity at that moment:

\[ v(1) = 2 + 10(1) + 9(1)^2 \]

Calculating this step by step:

1. \( 10(1) = 10 \)
2. \( 9(1)^2 = 9 \)

Now, substituting these values back into the equation:

\[ v(1) = 2 + 10 + 9 = 21 \]

### Step 3: Conclusion
The instantaneous velocity of the object at \( t = 1 \) second is:

\[ \boxed{21 \text{ m/s}} \]

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