Constant acceleration motion problem, please help!

  • Thread starter chroncile
  • Start date
  • #1
35
0

Homework Statement


An arrow is shot vertically upward with an initial speed of 25 m/s. Two seconds later, another arrow is shot upward with the same speed. On its way up, the second arrow meets the first arrow on its way down. Assume that both arrows experience a constant downward acceleration of 9.8 m/s2.

How long does it take (from the time the first arrow is shot) for the arrows to meet?


Homework Equations


d = vf * t - 0.5(a)(t)2
tA1 = tA2 + 2


The Attempt at a Solution


dA1 = 25t - 4.9t2
dA2 = 25t - 4.9t2

Since dA1 = dA2

25tA1 - 4.9tA12 = 25tA2 - 4.9tA22

Since tA1 = tA2 + 2

25(tA2 + 2) - 4.9(tA2 + 2)2 = 25tA2 - 4.9tA22

19.6t = 50-19.6
t = 1.55 s

The correct answer is 3.55 s
 

Answers and Replies

  • #2
berkeman
Mentor
59,971
10,171

Homework Statement


An arrow is shot vertically upward with an initial speed of 25 m/s. Two seconds later, another arrow is shot upward with the same speed. On its way up, the second arrow meets the first arrow on its way down. Assume that both arrows experience a constant downward acceleration of 9.8 m/s2.

How long does it take (from the time the first arrow is shot) for the arrows to meet?


Homework Equations


d = vf * t - 0.5(a)(t)2
tA1 = tA2 + 2


The Attempt at a Solution


dA1 = 25t - 4.9t2
dA2 = 25t - 4.9t2

Since dA1 = dA2

25tA1 - 4.9tA12 = 25tA2 - 4.9tA22

Since tA1 = tA2 + 2

25(tA2 + 2) - 4.9(tA2 + 2)2 = 25tA2 - 4.9tA22

19.6t = 50-19.6
t = 1.55 s

The correct answer is 3.55 s

I believe it's just that you have this equation off: tA1 = tA2 + 2

The 2nd arrow is shot 2 seconds after the first, not the other way around. Does that fix it?
 
  • #3
berkeman
Mentor
59,971
10,171
I believe it's just that you have this equation off: tA1 = tA2 + 2

The 2nd arrow is shot 2 seconds after the first, not the other way around. Does that fix it?

Actually, the only error is related to that equation. You forgot to add the 2 seconds to the time of the 2nd arrow being shot. You solved tor t2, and they asked for t1.
 
  • #4
35
0
I tried switching them around, but I still ended up with 1.55 s, help?
 
  • #5
berkeman
Mentor
59,971
10,171
I tried switching them around, but I still ended up with 1.55 s, help?

Did you see my last post about adding the 2 seconds?
 
  • #6
35
0
Can you please explain that?
 
  • #7
berkeman
Mentor
59,971
10,171
Can you please explain that?

The solution you got is for the time from firing the 2nd arrow up. But you fired the 1st arrow 2 seconds earlier. Therefore the time for the arrows to meet from the time you fired the first arrow is 1.55s + 2.0s = 3.55s.
 
  • #8
35
0
Okay, I get it now, thanks :smile:
 
  • #9
berkeman
Mentor
59,971
10,171
Okay, I get it now, thanks :smile:

You're welcome. You did a good job setting up and solving the problem. It makes it a lot easier to help when we get to see all of your work like that.
 

Related Threads on Constant acceleration motion problem, please help!

  • Last Post
Replies
9
Views
2K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
7
Views
2K
Replies
1
Views
12K
Replies
1
Views
919
Replies
1
Views
1K
Replies
3
Views
1K
Top