Please check my acceleration hockey answer.

  • Thread starter cbrowne
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  • #1
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Homework Statement



A hockey player is standing on his skates on a frozen pond when an opposing
player skates by with the puck, moving with a constant speed of 12 m/s. After
3.0 s, the first player makes up his mind to chase his opponent and starts
accelerating uniformly at 3.8 m/s^2.
(a) How long does it take him to catch his opponent?
(b) How far does he travel before he catches up with his opponent?


Homework Equations





The Attempt at a Solution



We know the initial velocities and accelerations of the two players,
a1 = 4 m/s2, a2 = 0
v1i = 0, v2i = 12 m/s
x1i = x2i = 0

The position of player #2 is given by
x2 = x2i + v2i t + (1/2) a2 t2
x2 = (12 m/s) t

Be careful with the time. We must account for player #1's wait of 3 s. With this accounted for, we can calculate the position of player #1 from
x1 = x1i + v1i (t - 3 s) + (1/2) a1 (t - 3 s)2

Of course, this equation only makes sense for t > 3 s.
x1 = 0 + 0 + (1/2) (3.8 m/s2) (t - 3 s)2 = (2 m/s2) (t2 - 6 s t + 9 s2)

Now we set x1 = x2 and solve for the time t.
x1 = x2
(1.9 m/s2) (t2 - 6 s t + 9 s2) = (12 m/s) t

We can either keep the units in explicitly or ensure that we have consistent units and simply write

1.9 (t2 - 6t + 9) = 12t
1.9 t2 - 12 t + 18 = 12 t
1.9 t2 - 24 t + 18 = 0

There are two solutions to this quadratic equation,
t1 = 11.8 s, and t2 = 0.8 s

However, the equation for the position of player #1 is not valid for t2 < 3 s, so we keep only tx,

t = 11.8 s

(b) How far has the first player traveled in this time?
Now where is player #2 (and, therefore, player #1 as well) at this time?
x2 = x2i + v2i t + (1/2) a2 t2
x2 = (12 m/s) t
x2 = (12 m/s) (11.8 s)
x2 = 141.6 m

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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It's really simpler if you start your time at the point that the second player starts to accelerate.

The player at rest is at x=0 at t=0.
At that moment player 1 is 36 meters ahead and moving at 12 m/s.

That makes it that when x for both is equal that will be at

1/2*3.8*t2 = 3*12 + 12*t

1.9*t2 - 12*t - 36 = 0

You know the quadratic formula from here. And either formula should yield the distance.
 
  • #3
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i get 2.21 if i use the quadratic equation for part b..
 
  • #4
LowlyPion
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i get 2.21 if i use the quadratic equation for part b..

When you have solved for t, you only have to plug that into x = 1/2*a*t2
 
  • #5
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o ok, so my answer of 141.6 was right
 
  • #6
LowlyPion
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o ok, so my answer of 141.6 was right

It's not what I get because you made a rounding simplification in yout equations.

This math is not correct:
1.9 (t2 - 6t + 9) = 12t
1.9 t2 - 12 t + 18 = 12 t
1.9 t2 - 24 t + 18 = 0

If you solve the quadratic for what I supplied you should get something like 8.53 sec.

Your time is based on the first skater's start time and is 3 sec longer than the correct answer in any event.
 
  • #7
30
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hmm.. when i plug
1.9 t2 - 24 t + 18 = 0

into the quad. eq i get 11.8308 or 0.800
 
  • #8
LowlyPion
Homework Helper
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hmm.. when i plug
1.9 t2 - 24 t + 18 = 0

into the quad. eq i get 11.8308 or 0.800

Yes and that is wrong from the statement of the problem.

Your time = 0 in your equations is 3 seconds before the skater actually starts. You must subtract 3 seconds from that.

But that aside you have not correctly reduced your own equation. Your quadratic is incorrect from the rounding that you introduced.
 
  • #9
30
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ok so then 11.8 - 3 = 8.8, and then that also solves the problem with the sig figs since the smallest number in my quadratic equation (1.9) has only 2 sig figs.
 
  • #10
LowlyPion
Homework Helper
3,090
5
ok so then 11.8 - 3 = 8.8, and then that also solves the problem with the sig figs since the smallest number in my quadratic equation (1.9) has only 2 sig figs.

Why don't you try solving your own equation correctly.
You have failed to correct your math error. You came up with :
1.9*t2 - 24*t + 18 = 0

If you had reduced your math correctly it would have been:
1.9*t2 - 23.4*t + 17.1 = 0

This yields the right answer after subtracting 3.
It bears a remarkable resemblance to my method I might note.
 
  • #11
30
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ok when i plug in 1.9*t2 - 23.4*t + 17.1 = 0

i get 11.83-3 =8.83--> sig figs = 8.8 seconds
 
  • #12
LowlyPion
Homework Helper
3,090
5
ok when i plug in 1.9*t2 - 23.4*t + 17.1 = 0

i get 11.83-3 =8.83--> sig figs = 8.8 seconds

Your answer is wrong still. You didn't solve the quadratic for that equation.
 
  • #13
30
0
really!? what exactly are you getting?
 

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