A vehicle is driven along a straight horizontal track by a motor which exerts a constant driving force. The vehicle starts from rest at `t=0` and the effects of friction and air resistance are negligible. If the kinetic eneryg of the vehicle at time t is E and power developed by the motor is P, which of the following graphs are correct?

Options

A. (a)
B. (b)
C. (c)
D. (d)

Explanation

To solve the problem, we need to analyze the motion of the vehicle under the influence of a constant driving force. We will derive the expressions for kinetic energy and power, and then determine the correct graphs for each.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The vehicle starts from rest, which means the initial velocity \( u = 0 \).
- The driving force \( F \) is constant, and since there are negligible effects of friction and air resistance, the vehicle will experience constant acceleration.

2. **Finding Acceleration**:
- According to Newton's second law, the acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} \]
- Here, \( m \) is the mass of the vehicle.

3. **Velocity as a Function of Time**:
- Since the vehicle starts from rest and accelerates uniformly, the velocity \( v \) at time \( t \) can be given by:
\[ v = u + at = 0 + \frac{F}{m} t = \frac{F}{m} t \]

4. **Kinetic Energy**:
- The kinetic energy \( E \) of the vehicle at time \( t \) is given by:
\[ E = \frac{1}{2} m v^2 \]
- Substituting the expression for \( v \):
\[ E = \frac{1}{2} m \left(\frac{F}{m} t\right)^2 = \frac{F^2}{2m} t^2 \]
- This shows that kinetic energy \( E \) varies with the square of time \( t^2 \), indicating a parabolic relationship.

5. **Power Developed by the Motor**:
- Power \( P \) is defined as the rate of doing work or the rate of change of energy. It can also be expressed as:
\[ P = F \cdot v \]
- Substituting the expression for \( v \):
\[ P = F \cdot \left(\frac{F}{m} t\right) = \frac{F^2}{m} t \]
- This indicates that power \( P \) varies linearly with time \( t \).

6. **Graph Analysis**:
- For kinetic energy \( E \), since it varies as \( t^2 \), the graph of \( E \) vs. \( t \) will be a parabola starting from the origin.
- For power \( P \), since it varies linearly with time, the graph of \( P \) vs. \( t \) will be a straight line starting from the origin.

### Conclusion:
- The correct graphs are:
- Kinetic Energy \( E \) vs. Time \( t \): Parabolic curve (option C).
- Power \( P \) vs. Time \( t \): Straight line (option A).