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.
The answer is 44.7, but I can't figure out how to get the same answer.
Figure the area under the acceleration-time plot. Just figure the area under the curve from t=-2 to t=6, then add the initial velocity at t=-2 of 10m/s, and you have it.
To solve this problem, we need to make use of the given information about the particle's acceleration and initial velocity. Here's the step-by-step approach to finding the particle's velocity at t = 6.0 s:
1. Examine figure 2-41 and note that the acceleration (a) is given along the vertical axis, and time (t) is given along the horizontal axis.
2. Find the time at which the particle's velocity is 10.0 m/s. From the graph, locate the point on the acceleration (a) axis where a = 13.0 m/s². Then, trace horizontally until you intersect with the line representing the velocity. From there, trace vertically until you reach the t = -2.0 s mark on the time (t) axis. This point represents the given initial velocity of 10.0 m/s.
3. Using the given initial velocity, calculate the total change in velocity from t = -2.0 s to t = 6.0 s. This can be done by finding the area under the velocity curve between these two time points.
4. Divide the region between t = -2.0 s and t = 6.0 s into two parts: the region below the x-axis and the region above the x-axis.
5. Calculate the area of the region below the x-axis separately. This negative area corresponds to a decrease in velocity.
6. Calculate the area of the region above the x-axis separately. This positive area corresponds to an increase in velocity.
7. Subtract the magnitude of the negative area from the positive area to find the net change in velocity.
8. Add the net change in velocity to the initial velocity (10.0 m/s) to obtain the final velocity at t = 6.0 s.
By following these steps, you should be able to calculate the final velocity and confirm if it matches the given answer of 44.7 m/s.