Splung.com Physics Forum Forum Index Splung.com Physics Forum
Discuss physics or ask questions. Part of the Splung.com physics web site.
 
   SearchSearch   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Physics Problem 2 - Distance, Time Problem

 
Post new topic   Reply to topic    Splung.com Physics Forum Forum Index -> Physics Help. Ask a Physics Expert
View previous topic :: View next topic  
Author Message
dumperdery



Joined: 23 May 2010
Posts: 4
Location: Canada

PostPosted: Mon May 24, 2010 5:04 pm    Post subject: Physics Problem 2 - Distance, Time Problem Reply with quote

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?
Back to top
View user's profile Send private message
hepcj
Site Admin


Joined: 23 Jun 2007
Posts: 125

PostPosted: Mon May 24, 2010 10:52 pm    Post subject: Reply with quote

This problem can be solved graphically or by writing equations of motion for the distance and time.
For the hockey player moving at constant speed, his equation of motion would be:
s(t) = 12 t
i.e. every second he moves 12 metres.
For the second hockey player, we need the equation of motion that gives the distance covered in terms of the acceleration as a function of time.
s(t) = u t + 0.5 a t^2
Since the second player does not move until 3 seconds after we need replace the time by t-3.
Then to find the time you need to equate the two equations and solve for t.
When you know t, you can substitute it back into either of the equations of motion to find the distance.

I make the time, 11.53 sec and the distance 138.44 metres.
Back to top
View user's profile Send private message Send e-mail
dumperdery



Joined: 23 May 2010
Posts: 4
Location: Canada

PostPosted: Tue May 25, 2010 8:49 pm    Post subject: Reply with quote

I got 8.53 seconds and 138 meters.

my equation was

(0.5)(3.Cool(T^2) = 12 (T + 3)

I solved for T using the quadratic formula and got 8.53 seconds. When you plug this into D = (Vi)(T) + (0.5)aT^2 you get 138 meters
Back to top
View user's profile Send private message
hepcj
Site Admin


Joined: 23 Jun 2007
Posts: 125

PostPosted: Tue May 25, 2010 10:26 pm    Post subject: Reply with quote

Y, it depend on when the time starts. 11.53 if the clock starts when the first hockey player starts to move. 8.53 seconds, if clock starts when the second player moves. Since the question says how long does it take for him to catch up the answer would be 8.53 secs.
Back to top
View user's profile Send private message Send e-mail
Display posts from previous:   
Post new topic   Reply to topic    Splung.com Physics Forum Forum Index -> Physics Help. Ask a Physics Expert All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


© Copyright 2010 Splung.com