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 can follow these steps:

### Step 1: Understand the relationship between velocity and acceleration
Acceleration \( a \) is defined as the rate of change of velocity with respect to time. Mathematically, this is expressed as:
\[ a = \frac{dv}{dt} \]

### Step 2: Differentiate the velocity function
Given the velocity function:
\[ v(t) = t^2 - 4t + 10^5 \]
we need to differentiate this function with respect to \( t \).

Using the power rule of differentiation:
- The derivative of \( t^2 \) is \( 2t \).
- The derivative of \( -4t \) is \( -4 \).
- The derivative of the constant \( 10^5 \) is \( 0 \).

Thus, the derivative \( \frac{dv}{dt} \) is:
\[ \frac{dv}{dt} = 2t - 4 \]

### Step 3: Substitute \( t = 1 \) second into the acceleration formula
Now, we substitute \( t = 1 \) into the expression for acceleration:
\[ a = 2(1) - 4 \]

### Step 4: Calculate the acceleration
Calculating the above expression:
\[ a = 2 - 4 = -2 \text{ m/s}^2 \]

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

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