The problem is from fundamentals of physics 9th edition by halliday.
Here is a free download of the book since I couldn't post the graph (I had to write out link since site doesn't allow links to be posted):
tripplew period sendspace period com/file/shrz0x
I am having trouble figuring out ch. 2 problem 82 (similar problem). My question is:
Figure 2-41
gives the acceleration a versus time t for a particle moving along an x axis. The a-axis scale is set by as = 13.0 m/s2. At t = -2.0 s, the particle's velocity is 10.0 m/s. What is its velocity at t = 6.0 s?
Note: Where it states as = 13.0, the s is supposed to be subscript.
To solve this problem, we need to analyze the graph given in Figure 2-41, which shows acceleration (a) versus time (t) for a particle moving along the x-axis.
First, let's understand what the graph is telling us. The x-axis represents time, and the y-axis represents acceleration. The graph shows how the acceleration of the particle changes over time.
Now let's address the specific question. At t = -2.0 s, the particle's velocity is 10.0 m/s. We want to find the velocity at t = 6.0 s.
To find the velocity at a specific time using the given graph, we need to integrate the acceleration-time graph to obtain the velocity-time graph. The area under the acceleration-time graph represents the change in velocity.
Since the acceleration-time graph is in the form of a curve, we can approximate the area under the curve by dividing it into smaller rectangles or trapezoids.
To do this, follow these steps:
1. Identify the region on the graph between t = -2.0 s and t = 6.0 s.
2. Divide this region into smaller regions such that each region represents a rectangle or a trapezoid.
3. Estimate the area of each of these smaller regions.
4. Add up all the areas of the smaller regions to calculate the total change in velocity.
5. Finally, add the initial velocity (10.0 m/s) to the calculated change in velocity to find the final velocity at t = 6.0 s.
Remember that the acceleration-time graph is given in terms of as = 13.0 m/s². Make sure to adjust your calculations accordingly using the correct scaling.
If you encounter any specific difficulties while attempting these calculations, please provide more details, and I can guide you through each step more precisely.