If `v=(t^(2)-4t+10^(5))`m/s where t is in second. Find acceleration at t=1 sec.

Options

A. 0
B. `2 m//s^(2)`
C. `-2 m//s^(2)`
D. None of these

Answer

To find the acceleration at \( t = 1 \) second given the velocity function \( v(t) = t^2 - 4t + 10^5 \) m/s, we will follow these steps:

### Step 1: Write down the velocity function
The velocity function is given as:
\[ v(t) = t^2 - 4t + 10^5 \]

### Step 2: Differentiate the velocity function to find acceleration
Acceleration \( a(t) \) is defined as the derivative of velocity with respect to time:
\[ a(t) = \frac{dv}{dt} \]
Now, we differentiate the velocity function:
\[ \frac{dv}{dt} = \frac{d}{dt}(t^2 - 4t + 10^5) \]
Using the power rule and the constant rule of differentiation:
- The derivative of \( t^2 \) is \( 2t \).
- The derivative of \( -4t \) is \( -4 \).
- The derivative of \( 10^5 \) (a constant) is \( 0 \).

Thus, we have:
\[ a(t) = 2t - 4 \]

### Step 3: Substitute \( t = 1 \) second into the acceleration function
Now, we substitute \( t = 1 \) into the acceleration function:
\[ a(1) = 2(1) - 4 \]
Calculating this gives:
\[ a(1) = 2 - 4 = -2 \text{ m/s}^2 \]

### Final Answer
The acceleration at \( t = 1 \) second is:
\[ \boxed{-2 \text{ m/s}^2} \]

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