Why Does Negative Initial Velocity Differ from the Textbook's Positive Value?

[Bubofthedead]

Homework Statement

What did I do wrong? My book's answer says the initial velocity when the rock was thrown straight down off a cliff was vi= 15.2m/s, but I got vi= -15.2m/s.

What is the initial speed of stone 2?

yi= 50m
yf= 0m
ti= 0s
tf=2.0s
ay= -9.8m/s^2
Vyf= 0m/s
Vyi= Unknown

Homework Equations

yf= yi + vi (change in time) + 1/2 (ay) (change in time)^2

1.)Subtract yi from both sides gives:

yf - yi = vi (change in time) + 1/2 (ay) (change in time)^2

2. Subtract (1/2 (ay) (change in time)^2) from both sides gives:

yf - yi - (1/2 (ay) (change in time)^2) = vi (change in time)

3. Divide both sides by (change in time) gives:

(yf - yi - (1/2 (ay) (change in time)^2)) / (change in time) = vi

The Attempt at a Solution

(yf - yi - (1/2 (ay) (change in time)^2)) / (change in time) = vi

(0m - 50m - (1/2 (-9.8m/s^2) (2.0s)^2)) / (2.0 s) = vi

= -15.2m/s

My textbook says vi= 15.2 m/s

 

[stewartcs]

Homework Statement

What did I do wrong? My book's answer says the initial velocity when the rock was thrown straight down off a cliff was vi= 15.2m/s, but I got vi= -15.2m/s.

What is the initial speed of stone 2?

yi= 50m
yf= 0m
ti= 0s
tf=2.0s
ay= -9.8m/s^2
Vyf= 0m/s
Vyi= Unknown

Homework Equations

yf= yi + vi (change in time) + 1/2 (ay) (change in time)^2

1.)Subtract yi from both sides gives:

yf - yi = vi (change in time) + 1/2 (ay) (change in time)^2

2. Subtract (1/2 (ay) (change in time)^2) from both sides gives:

yf - yi - (1/2 (ay) (change in time)^2) = vi (change in time)

3. Divide both sides by (change in time) gives:

(yf - yi - (1/2 (ay) (change in time)^2)) / (change in time) = vi

The Attempt at a Solution

(yf - yi - (1/2 (ay) (change in time)^2)) / (change in time) = vi

(0m - 50m - (1/2 (-9.8m/s^2) (2.0s)^2)) / (2.0 s) = vi

= -15.2m/s

My textbook says vi= 15.2 m/s

I believe the standard convention for g is 9.8 m/s^2, not -9.8 m/s^2. That would be my guess.

CS

 

[Borek]

Posting the question won't hurt... Very likely you are right, you are just mixing directions.

 

[Bubofthedead]

yeah, but the equation I stared from was linear horizontal motion, and I used it for linear vertical motion.

Xf = Xi + vi (change in time) + ½ a (change in time)^2

Yf = Yi + vi (change in time) + ½ ay (change in time)^2

where ay = afreefall= -g= -9.8m/s^2

Which should look like this

Yf = Yi + vi (change in time) + ½ (-g) (change in time)^2

Unless I am wrong

 

[Borek]

You have still not posted the question...

 

[Bubofthedead]

The full question says:

"A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws the two stones vertically downward 1.0s apart and observes that they cause a single splash. The initial speed of the first stone was 2.0m/s."

a. How long after the release of the first stone does the second stone hit the water?

b. What was the initial speed of the second stone?

c. What is the speed of each stone as they hit the water.


After taking the time to type the answer out, I noticed that the question mentions speed. Does this mean that the answer was negative if that asked for velocity, but positive if they asked for speed?

 

[Borek]

Never heard about difference between speed and velocity.

Acceleration works in the same direction stone was thrown.

 

[Redbelly98]

After taking the time to type the answer out, I noticed that the question mentions speed. Does this mean that the answer was negative if that asked for velocity, but positive if they asked for speed?

Yes. Assuming they took the upward direction as positive.

 

[Redbelly98]

Never heard about difference between speed and velocity.

In English-language physics: velocity is a vector, and speed is the magnitude of the velocity.

 

[Bubofthedead]

Thanks for the help

 

[Borek]

In English-language physics: velocity is a vector, and speed is the magnitude of the velocity.

Thanks. I thought they are just synonyms. None of the glossaries I have shows different meanings.

 

[Redbelly98]

If these are just English-language glossaries, they wouldn't. Scientific glossaries would (or should).

Did they make a similar distinction in your physics courses?

Thanks for the help

You're welcome.