| Topic: |
Science > Physics |
| User: |
"Jordan Taylor" |
| Date: |
02 Mar 2006 09:45:41 AM |
| Object: |
interesting kinematics problem |
Try:
a) http://tinyurl.com/egqvp
b) http://tinyurl.com/hn4bp (1.1, same as above if inaccessible)
1.1. A motorboat going downstream overcame a raft at point A; T=60
minutes later it turned back and after some time passed the raft at a
distance l=6.0 km from the point A. Find the flow velocity assuming the
duty of the engine to be constant.
No idea where I am going wrong.
.
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| User: "Greg Neill" |
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| Title: Re: interesting kinematics problem |
02 Mar 2006 10:01:47 AM |
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"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141314341.548949.214420@j33g2000cwa.googlegroups.com...
Try:
a) http://tinyurl.com/egqvp
b) http://tinyurl.com/hn4bp (1.1, same as above if inaccessible)
1.1. A motorboat going downstream overcame a raft at point A; T=60
minutes later it turned back and after some time passed the raft at a
distance l=6.0 km from the point A. Find the flow velocity assuming the
duty of the engine to be constant.
No idea where I am going wrong.
I can't see where you're going wrong either; you
haven't shown us what you've tried.
.
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| User: "Jordan Taylor" |
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| Title: Re: interesting kinematics problem |
02 Mar 2006 01:27:20 PM |
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(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
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| User: "Greg Neill" |
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| Title: Re: interesting kinematics problem |
02 Mar 2006 02:22:35 PM |
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"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
.
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| User: "Jordan Taylor" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 07:32:39 AM |
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Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
.
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| User: "Greg Neill" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 07:40:35 AM |
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"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141392759.443632.300850@i40g2000cwc.googlegroups.com...
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
You want to know the answer rather than how to
obtain the answer? What's the point?
.
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| User: "Jordan Taylor" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 11:39:03 AM |
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Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141392759.443632.300850@i40g2000cwc.googlegroups.com...
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
You want to know the answer rather than how to
obtain the answer? What's the point?
because I have the answer, just don't know how to obtain it. If your
method is correct, our answers should match.
.
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| User: "Greg Neill" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 12:47:42 PM |
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"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141407543.699077.290970@j33g2000cwa.googlegroups.com...
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141392759.443632.300850@i40g2000cwc.googlegroups.com...
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
You want to know the answer rather than how to
obtain the answer? What's the point?
because I have the answer, just don't know how to obtain it. If your
method is correct, our answers should match.
Give a man a fish and he will be fed for a day.
Teach a man to fish and he'll never be hungry.
So, do you want to fish or cut bait?
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| User: "PD" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 08:45:30 AM |
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Jordan Taylor wrote:
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
Hint: You are making this way, way harder than it is. You are looking
for the right equation to use and plug in numbers, but you should be
able to do it in your head in one mental line. One more little nudge:
Forget the motorboat. Just think about the two moments the raft was
spotted and where it was at those two moments.
PD
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| User: "Jordan Taylor" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 11:44:54 AM |
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PD wrote:
Jordan Taylor wrote:
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
Hint: You are making this way, way harder than it is. You are looking
for the right equation to use and plug in numbers, but you should be
able to do it in your head in one mental line. One more little nudge:
Forget the motorboat. Just think about the two moments the raft was
spotted and where it was at those two moments.
PD
I think I got it. 6.0km/v=(t+T) !!!!
gotcha!
.
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| User: "PD" |
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| Title: Re: interesting kinematics problem |
03 Mar 2006 01:11:09 PM |
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Jordan Taylor wrote:
PD wrote:
Jordan Taylor wrote:
Greg Neill wrote:
"Jordan Taylor" <Jordan.Taylor.beach@gmail.com> wrote in message
news:1141327640.534141.134730@i39g2000cwa.googlegroups.com...
(V + v)T - (V-v)t = 6.0km
v(T+t) = 6.0km
where V = speed of engine,
v = speed of flow
t=time taken to reach the raft after the motorboat turned,
T = 60 minutes
So you're stuck on determining what t is then.
If you ignore the flow of the river and imagine instead
that the raft and boat are on a placid lake so that
the raft remains fixed in the same place with respect
to the water, how long will the boat take to return to
the raft if it turns around after T=60 minutes?
Well, what is the answer you have for V?
Hint: You are making this way, way harder than it is. You are looking
for the right equation to use and plug in numbers, but you should be
able to do it in your head in one mental line. One more little nudge:
Forget the motorboat. Just think about the two moments the raft was
spotted and where it was at those two moments.
PD
I think I got it. 6.0km/v=(t+T) !!!!
gotcha!
I thought you were looking for a flow velocity, not a time (t+T). Still
stuck on formulas, I see. Note the raft is floating on the river, that
is it only goes as fast as the river flows.
Let me distract you with a relevant side-question. Suppose I am driving
my car and see you walking along the highway as I pass by you. I
continue driving, take my shirts to the cleaners, return a rental
video, go buy some pasta and some vegetables at the grocery, and then
60 minutes later as I'm driving on the same road, I see you again,
still walking along the highway but 6.0 km down the road from where I
first saw you. How fast are you walking?
Moral of the lesson: Do NOT go for formulas first. Draw it first.
Explain what's going on in words first.
PD
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