Very simple calculus problem graphs and velocity/time graphs to acceleration.

In summary, the conversation discusses a question about finding the average acceleration of a car, using a graph and data points. The person asking the question is aware of using the derivative to find acceleration, but is not allowed to fit a best line or derive an equation. They are seeking help on how to approach the problem using the concept of instantaneous and average change.
  • #1
TexasCow
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Homework Statement


I'm studying for a Calculus test and I am having trouble on a problem. And who better to ask than the members of physicsforums!?

Basically, we are studying average acceleration, velocity, secant and tangent lines. We are given the graph of a function, and its corresponding data points.

The question asks:
"Find the average acceleration of the car, in f/s, over the interval 0<t<50.(The inequality signs are acutally "less than or equal to", but I don't know how to input such characters.)

I'm aware that the derivative of a velocity time graph is its acceleration/time graph. So I assume that the slope of the line is the acceleration. Unfortunately, we are not permitted to fit a best line, or derive an equation. Is there another way of doing this? Thanks for any help.
 
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  • #2
Think about it this way: derivative is essentially the limit of dy/dx where "dy" stands for the change in y (e.g., from one data point to the next). Similarly for x. How would you apply dy/dx when the change isn't "infinitesimal" (i.e. limit)?

To recap:
Instantaneous change = derivative = we take the limit
Average change = ratio = we do not take the limit
 

1. What is a velocity/time graph?

A velocity/time graph is a graphical representation of an object's velocity over time. The slope of the graph represents the object's acceleration, with steeper slopes indicating a greater acceleration.

2. How do you calculate acceleration from a velocity/time graph?

The acceleration can be calculated by finding the slope of the velocity/time graph. This can be done by dividing the change in velocity by the change in time between two points on the graph.

3. What does a horizontal line on a velocity/time graph indicate?

A horizontal line on a velocity/time graph indicates that the object is not accelerating, as the velocity remains constant over time.

4. How does a simple calculus problem graph relate to velocity/time graphs?

A simple calculus problem graph can be used to determine the velocity or acceleration of an object at a specific point in time, which can then be represented on a velocity/time graph.

5. What is the relationship between acceleration and distance on a velocity/time graph?

The area under the curve on a velocity/time graph represents the distance travelled by an object. Therefore, the steeper the slope (higher acceleration), the greater the distance the object will cover in a given time period.

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