If velocity v varies with time t as `v=2t^(2)`, then the plot between v and `t^(2)` will be given as :

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the relationship between velocity \( v \) and \( t^2 \) given by the equation:

\[ v = 2t^2 \]

### Step 1: Identify the Variables
We have two variables in this equation:
- \( v \) (velocity) which is dependent on \( t^2 \)
- \( t^2 \) (the square of time) which is the independent variable

### Step 2: Rewrite the Equation
The equation can be rewritten in the form of \( y = mx + c \), where:
- \( y \) is the dependent variable (here, \( v \))
- \( x \) is the independent variable (here, \( t^2 \))
- \( m \) is the slope of the line
- \( c \) is the y-intercept

From the equation \( v = 2t^2 \), we can see that:
- \( y = v \)
- \( x = t^2 \)
- \( m = 2 \)
- \( c = 0 \)

### Step 3: Analyze the Graph
Since \( c = 0 \), the line will pass through the origin (0,0). The slope \( m = 2 \) indicates that for every unit increase in \( t^2 \), \( v \) increases by 2 units.

### Step 4: Plot the Graph
1. Draw the x-axis and y-axis.
2. Label the x-axis as \( t^2 \) and the y-axis as \( v \).
3. Since the slope is 2, for every increase of 1 in \( t^2 \), \( v \) will increase by 2. This means:
- When \( t^2 = 0 \), \( v = 0 \)
- When \( t^2 = 1 \), \( v = 2 \)
- When \( t^2 = 2 \), \( v = 4 \)
- When \( t^2 = 3 \), \( v = 6 \)
4. Plot these points and draw a straight line through them, starting from the origin.

### Conclusion
The plot between \( v \) and \( t^2 \) will be a straight line passing through the origin with a slope of 2.

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