If reation between distance and time is `s =a + bt + ct ^(2)` find initial velocity and acceleration.

Options

A. `b + 2ct, 2c`
B. `b, 2c`
C. `2c,b `
D. `b + 2c, 2c`

Answer

To solve the problem, we start with the given relation between distance (s) and time (t):

\[ s = a + bt + ct^2 \]

### Step 1: Find the expression for velocity
Velocity (v) is defined as the rate of change of distance with respect to time. Mathematically, this is expressed as:

\[ v = \frac{ds}{dt} \]

### Step 2: Differentiate the distance function
We differentiate the given equation \( s = a + bt + ct^2 \) with respect to time (t):

\[ \frac{ds}{dt} = \frac{d}{dt}(a + bt + ct^2) \]

Since \( a \) is a constant, its derivative is 0. The derivative of \( bt \) with respect to \( t \) is \( b \), and the derivative of \( ct^2 \) with respect to \( t \) is \( 2ct \). Thus, we have:

\[ v = 0 + b + 2ct = b + 2ct \]

### Step 3: Find the initial velocity
The initial velocity is the velocity at \( t = 0 \):

\[ v(0) = b + 2c(0) = b \]

So, the initial velocity \( v_0 \) is:

\[ v_0 = b \]

### Step 4: Find the expression for 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 5: Differentiate the velocity function
Now we differentiate the velocity function \( v = b + 2ct \) with respect to time (t):

\[ \frac{dv}{dt} = \frac{d}{dt}(b + 2ct) \]

Again, since \( b \) is a constant, its derivative is 0. The derivative of \( 2ct \) with respect to \( t \) is \( 2c \). Thus, we have:

\[ a = 0 + 2c = 2c \]

### Step 6: Conclusion
We have found the initial velocity and acceleration:

- Initial velocity \( v_0 = b \)
- Acceleration \( a = 2c \)

### Final Answer:
- Initial Velocity: \( b \)
- Acceleration: \( 2c \)

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