How to go from acceleration vs time graph to velocity vs time graph?

In summary: The acceleration is the slope of the velocity." So to find the velocity graph, you need to find the slope of the acceleration graph. In summary, using the graphical analysis method, you can determine the velocity and position of a motorcyclist over a given time period without using kinematic equations or functional evaluation of the integral. The slope of the acceleration graph represents the velocity and the area under the velocity graph represents the position.
  • #1
physicslover11
3
0

Homework Statement


The figure below shows the acceleration-versus-time graph for a motorcyclist riding along the x-axis. Assume that at t = 0, x = -1 m and v = 2 m/s.

Unless otherwise specified, use the graphical analysis to solve all the questions below. Do not use the kinematic equations or the functional evaluation of the integral, although, if you happen to know these methods, you may use them to check your graphical result.
1) Plot the velocity-versus-time graph for the first 40 s.
2) a) What is the position of the motorcyclist at t = 40 s? b) What is the distance traveled by the motorcyclist in the first 40 s?
3) a) What is the position of the motorcyclist at t = 15 s? b) Using the kinematic equations, find the position of the motorcyclist at t = 15 s. Compare your answer to that obtained in part a).
4) Roughly sketch the position-versus-time graph for the first 40 s.


Homework Equations


i know that Δv=aΔt, Δx=vΔt


The Attempt at a Solution


i don't know where i should start ! please help
 

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  • #2
You just find the slope of the velocity graph. Remember, acceleration is the SLOPE of velocity. So from t=0 to t=5, the acceleration is 2. So that means that the velocity graph looks like...?
 
  • #3
but how do i get the slope of the velocity graph if the velocity graph is what I am looking for?
 
  • #4
physicslover11 said:
but how do i get the slope of the velocity graph if the velocity graph is what I am looking for?

The value of the acceleration graph is the slope of the velocity graph. That's whatjohnqwertyful was saying.
 

Related to How to go from acceleration vs time graph to velocity vs time graph?

1. How do you calculate velocity from an acceleration vs time graph?

To calculate velocity from an acceleration vs time graph, you need to find the area under the curve. This can be done by dividing the graph into smaller sections, calculating the area of each section, and then summing them up. The resulting value will give you the average velocity for that time interval.

2. What does the slope of an acceleration vs time graph represent?

The slope of an acceleration vs time graph represents the rate of change of acceleration over time. A steeper slope indicates a higher rate of acceleration, while a flatter slope indicates a lower rate of acceleration.

3. Can you have a negative velocity on a velocity vs time graph?

Yes, negative velocity can be represented on a velocity vs time graph. It simply means that the object is moving in the opposite direction of the positive direction on the graph. For example, if a car is moving in the negative direction on a horizontal axis, it would have a negative velocity on a velocity vs time graph.

4. How do you find the initial velocity on a velocity vs time graph?

The initial velocity on a velocity vs time graph is represented by the y-intercept (where the line crosses the y-axis). If the line starts at the origin (0,0), then the initial velocity is 0. Otherwise, you can use the slope of the line and the time value at the starting point to calculate the initial velocity using the formula v = m/t (where m is the slope).

5. How can you tell if an object is accelerating or decelerating on a velocity vs time graph?

If the velocity vs time graph has a positive slope, then the object is accelerating. If the slope is negative, then the object is decelerating (or accelerating in the opposite direction). If the slope is 0, then the object is moving at a constant velocity (not accelerating or decelerating).

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