Velocity-time curve for a body projected vertically upwards is with times (s)

Options

A. `v=v_(0) e^(-S//R)`
B. `v=v_(0)e^(S//R)`
C. `v=v_(0) e^(-R//S)`
D. `v=v_(0) e^(R//S)`

Explanation

To solve the problem of finding the velocity-time curve for a body projected vertically upwards, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**: When a body is projected vertically upwards, it moves against the force of gravity. The acceleration due to gravity (g) is acting downwards, which we will take as negative in our calculations.

2. **Define Initial Conditions**:
- Let the initial velocity of the body at time \( t = 0 \) be \( v_0 \).
- The acceleration \( a \) can be defined as \( a = -g \) (where \( g \approx 10 \, \text{m/s}^2 \)).

3. **Use the Equation of Motion**:
- The relationship between velocity, acceleration, and time is given by:
\[ v = v_0 + at \]
- Substituting \( a = -g \):
\[ v = v_0 - gt \]

4. **Graphical Representation**:
- The equation \( v = v_0 - gt \) represents a straight line with a negative slope (-g) and a y-intercept at \( v_0 \).
- The x-axis represents time (t), and the y-axis represents velocity (v).

5. **Determine the Time of Flight**:
- The body will reach its maximum height when the velocity becomes zero. Setting \( v = 0 \):
\[ 0 = v_0 - gt \implies t = \frac{v_0}{g} \]
- This time \( t \) is the time taken to reach the maximum height.

6. **Complete the Graph**:
- The graph will start at \( (0, v_0) \) and will decrease linearly until it reaches the point \( (t, 0) \) where \( t = \frac{v_0}{g} \).
- After reaching the maximum height, the body will start falling back down, and the velocity will become negative as it moves downwards.

7. **Final Representation**:
- The velocity-time graph will be a straight line starting from \( v_0 \) at \( t = 0 \) and decreasing to zero at \( t = \frac{v_0}{g} \), then continuing into the negative velocity region as the body falls back down.

### Summary of the Graph:
- The graph is a straight line with a negative slope, starting from \( v_0 \) and crossing the time axis at \( t = \frac{v_0}{g} \).