The position-time relation of a particle moving along the x-axis is given by
`x=a-bt+ct^(2)`
where a, b and c are positive numbers. The velocity-time graph of the particle is

Options

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B.
C.
D.

Explanation

To find the velocity-time graph of the particle given the position-time relation \( x = a - bt + ct^2 \), we will follow these steps:

### Step 1: Differentiate the position function to find velocity
The velocity \( v \) is the first derivative of the position \( x \) with respect to time \( t \).

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

### Step 2: Apply the differentiation rules
- The derivative of a constant \( a \) is \( 0 \).
- The derivative of \( -bt \) is \( -b \).
- The derivative of \( ct^2 \) is \( 2ct \).

Thus, we have:

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

### Step 3: Write the velocity equation
The velocity as a function of time \( t \) is:

\[ v(t) = -b + 2ct \]

### Step 4: Analyze the velocity function
- At \( t = 0 \):
\[ v(0) = -b + 2c(0) = -b \]
This indicates that the initial velocity is negative since \( b \) is a positive number.

- To find when the velocity is zero, set \( v(t) = 0 \):
\[ 0 = -b + 2ct \implies 2ct = b \implies t = \frac{b}{2c} \]
This shows that the velocity becomes zero at a positive time \( t = \frac{b}{2c} \).

### Step 5: Determine the nature of the velocity-time graph
The equation \( v(t) = -b + 2ct \) is a linear equation in \( t \) with:
- A slope of \( 2c \) (which is positive since \( c \) is positive).
- A y-intercept of \( -b \) (which is negative).

### Step 6: Sketch the velocity-time graph
1. The graph starts below the time axis (negative velocity) at \( t = 0 \).
2. It crosses the time axis at \( t = \frac{b}{2c} \) (where velocity is zero).
3. After this point, the graph will continue to rise above the time axis (positive velocity) due to the positive slope.

### Conclusion
The velocity-time graph is a straight line that starts below the x-axis, crosses it at a positive time, and continues to rise.