Troubleshooting Answers to Easy Physics Questions on LON-CAPA

AI Thread Summary
The discussion revolves around troubleshooting answers for physics homework submitted through the LON-CAPA system. The user is confident in their answers for three questions but finds that the system does not accept them. After reviewing calculations, particularly for the time it takes for a ball to reach a certain height, it is determined that rounding errors may have affected the final answer. The consensus is that the system should accept answers that are close to the correct value, as minor discrepancies are common. The user expresses gratitude for the assistance received in resolving the issue.
Mert Kerem Yavuz
Messages
10
Reaction score
0
Member advised to use the homework template for posts in the homework sections of PF.
Hey guys, hello. There is a system belongs in internet called LON-CAPA, and my teacher told that homeworks are done via that system. So I logged in, the questions are not that difficult to solve. But system doesn't accept some of my answers which I'm so sure that being true. So I'm going to write 3 easy questions for you, I'm just wondering if I did a mistake or not. Here we go.

1 - The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.31 m/s3, B = 1.07 m/s2, C = -4.91 m/s, and D = 3.45 m.
Q-1 What is the acceleration of the object at t = 0.357 s?
(MY ANSWER IS 7,088 m/s^2)

2- A ball is thrown straight upward in the air at a speed of 17.1 m/s. Ignore air resistance.
Q-2.1 What is the maximum height the ball will reach?
(MY ANSWER IS 14,92m)

Q-2.2 How long will it take to reach 2.31 m above its initial position on the way down?
(MY ANSWER IS 3,34s)

Thank you guys for you answers already!
 
Physics news on Phys.org
Mert Kerem Yavuz said:
Hey guys, hello. There is a system belongs in internet called LON-CAPA, and my teacher told that homeworks are done via that system. So I logged in, the questions are not that difficult to solve. But system doesn't accept some of my answers which I'm so sure that being true. So I'm going to write 3 easy questions for you, I'm just wondering if I did a mistake or not. Here we go.

1 - The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.31 m/s3, B = 1.07 m/s2, C = -4.91 m/s, and D = 3.45 m.
Q-1 What is the acceleration of the object at t = 0.357 s?
(MY ANSWER IS 7,088 m/s^2)

2- A ball is thrown straight upward in the air at a speed of 17.1 m/s. Ignore air resistance.
Q-2.1 What is the maximum height the ball will reach?
(MY ANSWER IS 14,92m)

Q-2.2 How long will it take to reach 2.31 m above its initial position on the way down?
(MY ANSWER IS 3,34s)

Thank you guys for you answers already!
I agree with your first two answers but not the third. Please post your working.
 
haruspex said:
I agree with your first two answers but not the third. Please post your working.

I solved this in two steps. First, I calculated the landing time.

I calculated g=9.8 m/s^2.
V= Vo+g*t
0=17.1+(9.8)*t
tlanding=1.744 s (landing time)

Then, i calculated the time between the maximum height position to 2,31 m above ground position.maximum height=14,92 m (Found it another question, which you said that you found it too)
14.92-2.31=12.61 m to go down
X=Vo*t+(1/2)*g*t^2
12.61=0+(1/2)*9.8*t^2
t=1.6 s

t1+t2=3.34 s (whole process)
 
Mert Kerem Yavuz said:
I solved this in two steps. First, I calculated the landing time.

I calculated g=9.8 m/s^2.
V= Vo+g*t
0=17.1+(9.8)*t
tlanding=1.744 s (landing time)

Then, i calculated the time between the maximum height position to 2,31 m above ground position.maximum height=14,92 m (Found it another question, which you said that you found it too)
14.92-2.31=12.61 m to go down
X=Vo*t+(1/2)*g*t^2
12.61=0+(1/2)*9.8*t^2
t=1.6 s

t1+t2=3.34 s (whole process)
Ok, I found my mistake, but it is nearer to 3.35. You may have got a bit of rounding error because you calculated two stages and added them together. That is not necessary. You can solve it in one step with s=ut+1/2 at2. You just have to pick the larger root of the quadratic.
 
haruspex said:
Ok, I found my mistake, but it is nearer to 3.35. You may have got a bit of rounding error because you calculated two stages and added them together. That is not necessary. You can solve it in one step with s=ut+1/2 at2. You just have to pick the larger root of the quadratic.

So in conclusion, you tell me that you found around 3.34,-3.35. So system must accept my answers cause the answer don't have to be the absolute true. In my other calculations, it accepted answers like 7,1 which's true answer is 7,15.
 
[Mod note: Please note that I have changed the thread title to better reflect the thread content. Pleas for help are not descriptive of the physics to be discussed and not allowed as thread titles]
 
Mert Kerem Yavuz said:
So in conclusion, you tell me that you found around 3.34,-3.35. So system must accept my answers cause the answer don't have to be the absolute true. In my other calculations, it accepted answers like 7,1 which's true answer is 7,15.
Yes.
 
haruspex said:
Yes.
Thanks, really appreciate it.
 
Back
Top