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Old Oct 12th 2017, 08:57 AM   #1
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Velocity time graph

Really stuck on this...

The velocity time graph of a diver who springs off of a board, up to a max height, and then down where he enters the pool and then swims through the water until he pushes off the floor of the swimming pool and then reaches the surface.

The bit where he springs off the board is presumably a constant negative acceleration to zero and then a change in direction so increasing velocity until he hits the pool. But then I am confused...
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Old Oct 12th 2017, 09:35 AM   #2
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initial positive velocity;
negative acceleration till zero velocity;
the same negative acceleration until hits water;
change to a positive acceleration (due to buoyancy) until hits floor;
sudden change in velocity, but the same buoyancy acceleration until surface.
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Old Oct 12th 2017, 09:46 AM   #3
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Originally Posted by Woody View Post
initial positive velocity;
negative acceleration till zero velocity;
the same negative acceleration until hits water;
change to a positive acceleration (due to buoyancy) until hits floor;
sudden change in velocity, but the same buoyancy acceleration until surface.
After he hits the floor of the pool, his velocity suddenly changes, is it now positive ( if it was negative before).
In other words, the second half of the graph is it beneath the axes or above?
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Old Oct 12th 2017, 09:52 AM   #4
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Are either of these correct?

Are either of these correct?

Velocity time graph-img_0568.jpg

Sorry needs rotating

Last edited by rodders2357; Oct 12th 2017 at 09:53 AM. Reason: sorry needs rotating!
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Old Oct 12th 2017, 03:08 PM   #5
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Try drawing 2 graphs.

First showing the acceleration against time
basically 2 flat lines at various levels,
one being the acceleration due to gravity (while in the air),
the other being the acceleration due to buoyancy (while in the water).

There is the additional feature of the push off the bottom of the pool,
This would be a large upward acceleration but for a very short time.

Now draw the velocity graph under the acceleration graph,
The level of the acceleration line at any time shows the slope of the velocity line at that time.
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Old Oct 12th 2017, 11:06 PM   #6
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Thanks Woody,

thats helpful. The thing i am not clear about is what direction is acceleration is for the three sections:
- as the diver is travelling through the water to the bottom
- the sudden upward acceleration.
- the upward section of the journey

is it positive, negative, negative?
just like dropping a ball? or does the buoyancy change things?
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Old Oct 16th 2017, 10:39 AM   #7
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Sorry about the delay

When in the air the acceleration is that due to gravity (down)
when in the water the acceleration will be due to buoyancy (up)

There are a couple of other time points which cause a velocity change,
one at the beginning (jumping off the board) and one at the bottom of the pool (pushing off).

One could treat these as short duration high accelerations (both upward)
but the question seems to be posed in such a way as to suggest that these two points should be treated as points of "instant" velocity change.
This is a reasonable approximation if the acceleration is high and the duration is short (as it will be for both these points).

So the diver starts with a positive velocity (imparted by the spring board),
this decreases (linearly) to zero, at the top of his jump
it continues to decrease (at the same linear rate) into negative velocities, until he hits the water.
The velocity starts to increase (become less negative) due to buoyancy,
but is still slightly negative when he reaches the bottom of the pool.
he then pushes off, switching suddenly from a small negative to a large positive velocity.
his velocity will then increase (due to buoyancy) until he reaches the surface.
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Last edited by Woody; Oct 16th 2017 at 10:51 AM.
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