Velocity graph

Well, this is really confusing me. I am suppose to be graphing a given situation into a v(t) vertical velocity graph.

The following occurs in 6 seconds (6 points on the graph):
Starting at y = 0 m, a stone is thrown straight up on the edge of a cliff until it reaches its maximum height at 2 seconds, and begins to fall. At 3 seconds, it is level with the 1 second point. At 4 seconds, the stone is level with y = 0 m, where it continues downward for another 2 seconds, ending at the sixth second. My displacement graph was correct, but my velocity graph is really confusing. In particular I am having trouble finding how exactly to plot this.

Things I figured out, but don't know if it's correct:

a. Going up 2 seconds is the same as free falling for 2 seconds? So the initial velocity is +19.6 m/s.
b. At the 1 second mark, the velocity is the (free-fall value of t=2) - (free-fall value of t=1) = +14.7 m/s
--------------------------------------------
c. The velocity at the 2 second mark is 0.
d. At the 3 second mark, the velocity is the free-fall value of t=1. So 3 = -4.9 m/s.
e. At the 4 second mark, the velocity is the (free-fall value of t=2)/2. -9.8 m/s
f. At 5 seconds, (free-fall value of t=3)/3. v = -14.7 m/s
g. At 6 seconds, (free-fall value of t=4)/4. v = -19.6 m/s

I know that something or everything is wrong, but I can't seem to think straight at the moment.
 
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Velocity is a vector so graphing it in terms of t will give you a vector field which is not what you want I think. You should just graph the speed of the object.
 
I have a predefined graph, and it only goes to -25, and +25 on the y axis (which is in meters). I initially was doing speed, but it doesn't fit (it's suppose to fit).

The following hints were given:
"The acceleration of the stone (due to gravity) is constant. Therefore, the slope of the velocity vs. time graph is constant."
Which I am getting at least on the tail end, but not sure if the numbers are right either.

"The velocity of the stone becomes negative after it reach its highest point."

I'm not even sure I was even finding the velocity before, or just incorrectly manipulating the numbers to make it fit.
 

Andrew Mason

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Thunderer said:
a. Going up 2 seconds is the same as free falling for 2 seconds? So the initial velocity is +19.6 m/s.
Ok.
b. At the 1 second mark, the velocity is the (free-fall value of t=2) - (free-fall value of t=1) = +14.7 m/s
How do you get velocity at t=1 of 14.7 m/sec? This is where your mistake is. What is the speed after 1 second if accelerating at -9.8m/sec and an initial speed of 19.6 m/sec?

Use:
[tex]v = v_0 + at [/tex] where a = g = -9.8m/sec^2

AM
 
19.6 m/s + (-9.8 m/s^2)(1 s)

That makes it 9.8 m/s right? It will continue the same trend as well...
0: 19.6 m/s
1: 9.8 m/s
2: 0 m/s
3: -9.8 m/s
4: -19.6 m/s
5: -29.4 m/s
6: -39.2 m/s

Is that right?

There's actually no possible way to plot all 6 velocities I'm guessing.
 

Andrew Mason

Science Advisor
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Thunderer said:
There's actually no possible way to plot all 6 velocities I'm guessing.
?? What makes it impossible to plot the function v(t) vs. t where [itex]v(t) = v_0 + gt[/itex]

AM
 
e(ho0n3 said:
Velocity is a vector so graphing it in terms of t will give you a vector field which is not what you want I think. You should just graph the speed of the object.
Sounds right...
 

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