If `v=(t+2)(t+3)` then acceleration`(i.e(dv)/(dt))` at t=1 sec.

Options

A. `5 m//s^(2)`
B. `7 m//s^(2)`
C. `2 m//s^(2)`
D. None of these

Answer

To find the acceleration given the velocity function \( v = (t + 2)(t + 3) \), we will follow these steps:

### Step 1: Expand the velocity function
First, we need to expand the expression for velocity:

\[ v = (t + 2)(t + 3) \]

Using the distributive property (FOIL method):

\[ v = t^2 + 3t + 2t + 6 = t^2 + 5t + 6 \]

### Step 2: Differentiate the velocity function
Next, we need to find the acceleration, which is the derivative of the velocity with respect to time \( t \):

\[ a = \frac{dv}{dt} = \frac{d}{dt}(t^2 + 5t + 6) \]

Using the power rule for differentiation:

\[ \frac{d}{dt}(t^n) = n \cdot t^{n-1} \]

We differentiate each term:

- The derivative of \( t^2 \) is \( 2t \).
- The derivative of \( 5t \) is \( 5 \).
- The derivative of the constant \( 6 \) is \( 0 \).

Thus, we have:

\[ a = 2t + 5 \]

### Step 3: Substitute \( t = 1 \) second into the acceleration equation
Now, we will find the acceleration at \( t = 1 \) second:

\[ a = 2(1) + 5 \]

Calculating this gives:

\[ a = 2 + 5 = 7 \, \text{m/s}^2 \]

### Final Answer
The acceleration at \( t = 1 \) second is \( 7 \, \text{m/s}^2 \).

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