#13 is a pretty long problem. You need to calculate the displacement for each of the three sections. They give you enough information, but each section is a separate calculation.
If you get stuck on one part, post a more specific question so we can give more specific help. Good luck.
I am also having trouble with this problem. I need some help with the 3rd part of the displacement.(When the acceleration is less than zero for x amount of seconds.
First you figure out the displacement for step 1. Then you figure out the displacement for step 2, and for that step you also need the velocity for step 2 (which is the final velocity for step 1).
Since the velocity doesn't change during step 2, the initial velocity for step 3 is the velocity you used for step 2. They give you the acceleration for step 3, and yes, it is a negative number. So for step 3 you have an initial velocity, an acceleration, and a time, and you are looking for displacement. With your information you should be able to find the displacement.
Good luck, and remember to keep lots of significant figures for all your calculations.
So close ... so close... It's a little hard to tell what you did since you didn't include units, but let me see if I can follow.
In step 1 you used the x= equation, and since xo=0 and vo=0 the equation becomes displacement x=(1/2)at^2. That should be right.
For step 2 you needed to find the final velocity from step 1 (the ending velocity from step 1 is the starting velocity for step 2). It looks as though you used the equation for average velocity v(bar)=(Δx)/t. That gets you the average velocity for step 1, and since it had started at rest (vo=0), the average velocity and the final velocity are different. Think: if you were in a car and you went from 0 to 60mph. The average would be 30mph (average), not 60mph (final). I think if you find the final velocity for step 1 (not the average velocity) then you will be on the right track. Everything else looks good.
hi Dr. Winters, I found the answer to question 13 but I am having trouble finding the answer to question 14, which has to do with leg 1 from #13. Question 14 asks me for the average speed of leg one. How would I find the average speed if for leg one I am only given a number for the time for the acceleration? I thought I had to use the average velocity formula to find the average speed but I don't know how to do this when I am only given acceleration and time.
You are probably over thinking this question. In #13 you found the distance (same as the displacement since it was in a straight line) for the first leg of the trip, and they give you the time for the first leg of the trip. From that information you should be able to use the average speed equation to find the average speed for the first leg of the trip, which is the answer for #14. Good luck.
8 comments:
#13 is a pretty long problem. You need to calculate the displacement for each of the three sections. They give you enough information, but each section is a separate calculation.
If you get stuck on one part, post a more specific question so we can give more specific help.
Good luck.
I am also having trouble with this problem. I need some help with the 3rd part of the displacement.(When the acceleration is less than zero for x amount of seconds.
First you figure out the displacement for step 1. Then you figure out the displacement for step 2, and for that step you also need the velocity for step 2 (which is the final velocity for step 1).
Since the velocity doesn't change during step 2, the initial velocity for step 3 is the velocity you used for step 2. They give you the acceleration for step 3, and yes, it is a negative number. So for step 3 you have an initial velocity, an acceleration, and a time, and you are looking for displacement. With your information you should be able to find the displacement.
Good luck, and remember to keep lots of significant figures for all your calculations.
# 13 For leg 1;
x=(1/2) * 2.88 * 15.4 *15.4=341.51 m
velocity= 341.41 / 15.4=22.176 m/s2
leg2 ; x= 22.176 * 225s = 4989.6 m
leg 3; x= 22.176 * 3.27 + (½) * (-8.74) * 3.27 * 3,27 = 25.7
total displacement = 341.51 + 4989.6 + 25.7=5356.897947 m/1000=5.356897947 km
Why is it wrong?
So close ... so close... It's a little hard to tell what you did since you didn't include units, but let me see if I can follow.
In step 1 you used the x= equation, and since xo=0 and vo=0 the equation becomes displacement x=(1/2)at^2. That should be right.
For step 2 you needed to find the final velocity from step 1 (the ending velocity from step 1 is the starting velocity for step 2). It looks as though you used the equation for average velocity v(bar)=(Δx)/t. That gets you the average velocity for step 1, and since it had started at rest (vo=0), the average velocity and the final velocity are different. Think: if you were in a car and you went from 0 to 60mph. The average would be 30mph (average), not 60mph (final). I think if you find the final velocity for step 1 (not the average velocity) then you will be on the right track. Everything else looks good.
Good luck.
hi Dr. Winters, I found the answer to question 13 but I am having trouble finding the answer to question 14, which has to do with leg 1 from #13. Question 14 asks me for the average speed of leg one. How would I find the average speed if for leg one I am only given a number for the time for the acceleration? I thought I had to use the average velocity formula to find the average speed but I don't know how to do this when I am only given acceleration and time.
You are probably over thinking this question. In #13 you found the distance (same as the displacement since it was in a straight line) for the first leg of the trip, and they give you the time for the first leg of the trip. From that information you should be able to use the average speed equation to find the average speed for the first leg of the trip, which is the answer for #14.
Good luck.
Yes thank you I got it!
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