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

Options

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

Answer

To solve the problem where the relationship between distance and time is given by the equation \( s = a + bt + ct^2 \), we need to find the initial velocity and acceleration. Let's break this down step by step.

### Step 1: Understand the given equation
The equation \( s = a + bt + ct^2 \) represents the displacement \( s \) as a function of time \( t \), where:
- \( a \) is a constant,
- \( b \) is the coefficient of \( t \),
- \( c \) is the coefficient of \( t^2 \).

### Step 2: Find the initial velocity
The initial velocity \( v_0 \) can be found by differentiating the displacement \( s \) with respect to time \( t \). The velocity \( v \) is given by:
\[ v = \frac{ds}{dt} \]
Differentiating \( s = a + bt + ct^2 \):
\[ \frac{ds}{dt} = \frac{d}{dt}(a) + \frac{d}{dt}(bt) + \frac{d}{dt}(ct^2) \]
Since \( a \) is a constant, its derivative is 0. The derivative of \( bt \) is \( b \), and the derivative of \( ct^2 \) is \( 2ct \). Therefore:
\[ v = 0 + b + 2ct = b + 2ct \]
To find the initial velocity, we evaluate this expression at \( t = 0 \):
\[ v_0 = b + 2c(0) = b \]
Thus, the initial velocity \( v_0 \) is \( b \).

### Step 3: Find the acceleration
Acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
We already found the expression for velocity \( v = b + 2ct \). Now, we differentiate this with respect to \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(b) + \frac{d}{dt}(2ct) \]
The derivative of \( b \) (a constant) is 0, and the derivative of \( 2ct \) is \( 2c \). Therefore:
\[ a = 0 + 2c = 2c \]
Thus, the acceleration is \( 2c \).

### Summary of Results
- The initial velocity \( v_0 \) is \( b \).
- The acceleration \( a \) is \( 2c \).