I need help with this ## Understanding Velocity-Time Graphs### OverviewThe area under the velocity vs time graph represents the distance traveled along a straight line. Here's how key concepts relate:- **Displacement**: Final position minus initial position.- **Distance**: The actual path traveled.**Formula:**- Average velocity = (displacement along a straight line) / (given time interval), or \( v = \frac{d}{t} \) and \( d = vt \).### Simulation InstructionsSet the initial position to \(-10 \, \text{m}\), velocity to \(5 \, \text{m/s}\), and acceleration to \(0.0\). Run the simulation until the object hits the wall. Calculate the area under the velocity curve when the object reaches the \(+10 \, \text{m}\) mark. This involves calculating the area between the velocity curve and the horizontal time axis.### Simulation Diagram DescriptionThe diagram displays a simulation with time on the \(x\)-axis and velocity on the \(y\)-axis. The graph shows a horizontal line indicating constant velocity. The objective is to determine the distance by calculating the area under this line.### Questionsa. What was the distance covered? _______ mb. What is the area under the velocity curve? _______ mc. What was the displacement of the man? _______ md. What was the speed of the man? _______ m/s (Remember speed = distance traveled / given time interval)e. What was the velocity? _______ m/s and direction (if any) _______ (Remember velocity = displacement / given time interval)

Required : Area under the velocity curve.

12. The area under the velocity vs time graph is supposed to be equal to the distance traveled along a straight line. Displacement = final position - initial position Distance: the distance traveled along the actual path. Since the average velocity = displacement (distance traveled along a straight line) / given time interval v = d/t d = vt Set the position of the man at -10.0 m, the velocity equal to 5 m's and the acceleration to 0.0. Run the simulation until the Moving Man hits the wall. What is the area under the velocity curve when the man reaches the +10 m mark? (You want velocity curve and the horizontal time axis. Realizing what kind of shape it is may help you calculate the area). Insert below a screen shot of the simulation for above description and your worksheet which shows how you obtain final answer for each question 39 seconds

12. The area under the velocity vs time graph is supposed to be equal to the distance traveled along a straight line. Displacement = final position - initial position Distance: the distance traveled along the actual path. Since the average velocity = displacement (distance traveled along a straight line) / given time interval v = d/t d = vt Set the position of the man at -10.0 m, the velocity equal to 5 m's and the acceleration to 0.0. Run the simulation until the Moving Man hits the wall. What is the area under the velocity curve when the man reaches the +10 m mark? (You want velocity curve and the horizontal time axis. Realizing what kind of shape it is may help you calculate the area). Insert below a screen shot of the simulation for above description and your worksheet which shows how you obtain final answer for each question 39 seconds