What is the Time and Distance for a Hockey Player to Catch Up with an Opponent?

[cbrowne]

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/s2. (a) How long does it take him to catch his
opponent? (b) How far does he travel before he catches up with his opponent?

The Attempt at a Solution

The position of player #2 is given by

x2 = x2i + v2i t + (1/2) a2 t2

x2 = (12 m/s) t

Equation: x1 = x1i + v1i (t - 3 s) + (1/2) a1 (t - 3 s)2


x1 = 0 + 0 + (1/2) (3.8 m/s2) (t - 3 s)2 = (1.9 m/s2) (t2 - 6 s t + 9 s2)

x1= x2

1.9t²- 11.4t+17.1=12t
1.9t²-23.4t+17.1=0

Now we plug these numbers into the quadratic equation and we get:

11.53 seconds

So the answer for PART A is 11.53 seconds.

b)
x2 = x2i + v2i t + (1/2) a2 t2

x2 = (12 m/s) t

x2 = (12 m/s) (11.53 s)

x2 = 138.36 m

The answer for PART b is 138.36m

 

[alphysicist]

Hi cbrowne,

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/s2. (a) How long does it take him to catch his
opponent? (b) How far does he travel before he catches up with his opponent?

The Attempt at a Solution

The position of player #2 is given by

x2 = x2i + v2i t + (1/2) a2 t2

x2 = (12 m/s) t

Equation: x1 = x1i + v1i (t - 3 s) + (1/2) a1 (t - 3 s)2


x1 = 0 + 0 + (1/2) (3.8 m/s2) (t - 3 s)2 = (1.9 m/s2) (t2 - 6 s t + 9 s2)

x1= x2

1.9t²- 11.4t+17.1=12t
1.9t²-23.4t+17.1=0

Now we plug these numbers into the quadratic equation and we get:

11.53 seconds

So the answer for PART A is 11.53 seconds.

I don't think this is correct. I believe they want to know how long the first player was in motion before he caught the second player. That is not what t represents in your equations.

 

[cbrowne]

thanks for replying. I have no idea what to do or where to start for this question. I was baseing my answer from this website: http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch02/Ch2.html

if you scroll to the bottom its pretty much the same question but instead of 3.8 m/s its 4.0 m/s.

Any help would be greatly appreciated.

 

[alphysicist]

The question is when do you start the clock? When they say how long does it take for the first person to catch the second, do they mean how long from the time the second person passed the first, or how long from the time the first person began chasing him?

From the way the question is worded, I think it means that they want to know how long the first person was moving; what you found was how long the second person was moving (after he passed the first). There is a three second difference between those times.