Physics Question

OK, I have about reached the end of my wits. But there's a ton of smart people on here... so I figured I should ask here.

I'm taking AP Physics. We just did a lab in which we rolled a ball down a ramp and timed it to determine its velocity.

Anyway, if you calculate distance traveled/time, you should get velocity, right?

According to our book, that is wrong. It says that velocity=(2*distance)/time. Which doesn't make sense. That would mean if I am traveling 60 miles an hour, I only go 30 miles every hour (which is about the biggest contradiction I've ever heard).

Does anyone have any thoughts about this? At first I thought it was a typo but then the teacher says this in the lab handout:
Question 1. If velocity is calculated distance/time, why are you being asked to calculate it using 2 times distance/time?

Which doesn't make sense....

Please help me retain my mental sanity.

Sounds like some kind of Common Core thing. Just talk about how it makes you feel, and you'll get it right.

In all seriousness, that makes no sense. V = d / t, with the distinction being that velocity is a vector, i.e., movement in a specific direction. You could be running back and forth between points A and B on line at a speed of 10 mph, but have a velocity of 0, since you were going backwards half the time. Or, if I am on point A trying to get to point B, which is ahead of me, and I drive backwards at 30 mph, then my speed is 30 mph, but my velocity is -30 mph, because I went backward.

No clue what is intended by doubling distance. I'd have to see the book.

blackbrandt said:

OK, I have about reached the end of my wits. But there's a ton of smart people on here... so I figured I should ask here.

I'm taking AP Physics. We just did a lab in which we rolled a ball down a ramp and timed it to determine its velocity.

Anyway, if you calculate distance traveled/time, you should get velocity, right?

According to our book, that is wrong. It says that velocity=(2*distance)/time. Which doesn't make sense. That would mean if I am traveling 60 miles an hour, I only go 30 miles every hour (which is about the biggest contradiction I've ever heard).

Does anyone have any thoughts about this? At first I thought it was a typo but then the teacher says this in the lab handout:
Question 1. If velocity is calculated distance/time, why are you being asked to calculate it using 2 times distance/time?

Which doesn't make sense....

Please help me retain my mental sanity.

Click to expand...

Your formula gives average velocity over the distance traveled. The book is giving you velocity at the end.
The assumption is that acceleration (a) is constant. So

V=a*time

Integrate that WRT time and you get

Distance = .5*a*Time^2

Substitute the first formula into the second:

Distance = .5*V*Time (Because the ball is speeding up as it rolls)

Transpose to get the book formula.

Again, your own formula gives the average speed, not the maximum.

Hope that helps

It depends on which velocity you are measuring. starting/ending or average.

You can define distance under constant acceleration as d = t*(u+v)/2 where u = starting velocity and v = ending velocity.

substituting u = 0 and rearranging gives you an ending velocity v = (2*d)/t


Edit: Larry beat me to it.

Very good explanation, to add to this, if you had an initial velocity, this would include
V= Vinit + aconst*time
then the integral with respect to time would give distance as a function of time
d= Vinit*time + 0.5*aconst*time^2
if you were going at a constant 60mph, then acceleration would be zero and you get the formula that you were expecting.

Vinit is zero at the top of the ramp and you have a constant accel due to gravity so you get only the second term. This gives the final velocity at the end of the measured time interval from the top start to the bottom of the ramp. This is not the average velocity over that time.
d=0.5*aconst*time^2
Vfinal=(aconst*time)
d=0.5*(Vfinal)*time
Vfinal=2*d/t

the average velocity would be (Vinit+Vfinal)/2 = (0 + 2*d/t)/2 = d/t = total distance/total time = avg vel
this avg vel equation (Vinit+Vfinal)/2 only works when accel is a constant value.

You guys all pass, and I flunk. :bang:

It's possible that your teacher is looking for you to figure velocity based on an Energy balance equation, taking the potential energy of the ball at the top of the ramp and converting to kinetic energy at the bottom (minus friction and air resistance losses). But you would have needed to measure the height of the ramp and mass of the ball during the lab.

Nathan said:

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Thanks for giving me bad college flashbacks.

Nathan said:

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...somehow it all makes sense now...

That's quantum mechanics which means now we can't simultaneously know both the position and the momentum of the ball on the ramp.

Swissyhawk said:

That's quantum mechanics which means now we can't simultaneously know both the position and the momentum of the ball on the ramp.

Click to expand...

Your ball is now in a alternate universe. No ball=no problem. Your answer is zero......

dave carver said:

...somehow it all makes sense now...

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Yeah. The Dirac notation puts it all into bold relief.

Arrggghhh! What happened to the epoxy thread? Yea, I said it!

No kidding. Which reminds me. I gotta check my RockerPoxy for Nute... Now, if I can just reach it from my recliner....

Adrian

One
and
one
and
one
is
three.

Unless one and one and one equals A, then all bets are off.

I notice BB hasn't returned to this thread - maybe we scared him off.

RandyT0001 said:

One
and
one
and
one
is
three.

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Hmmmm......where did you learn your chipher'n?

Everyone knows 1+1+1 equals 111!

Swissyhawk said:

That's quantum mechanics which means now we can't simultaneously know both the position and the momentum of the ball on the ramp.

Click to expand...

Since the ball has a fixed mass m and momentum M is defined by M=mv, this means that finding the velocity at any point on the ramp is impossible. Have a few brews with your professor and talk it over.

Zebedee said:

I notice BB hasn't returned to this thread - maybe we scared him off.

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Yeah... that $%^& load of math scared me off :rofl:

blackjack2564 said:

Hmmmm......where did you learn your chipher'n?

Everyone knows 1+1+1 equals 111!

Click to expand...

1+1+1=11 (binary)

blackjack2564 said:

Hmmmm......where did you learn your chipher'n?

Everyone knows 1+1+1 equals 111!

Click to expand...

jeff2space said:

1+1+1=11 (binary)

Click to expand...

He roller coaster
Got early Warning
Got muddy water
He mojo filter
Got to be good lookin'
He so hard to see
Come together
Right now
Over me

NAR29996 said:

Thanks for giving me bad college flashbacks.

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Dang, he took all the fun. Nothing left to do but plug and chug.

Suddenly,

I have a headache.

You all have beautiful minds.

Nathan said:

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Oh gods! P-Chem I!!
*Runs screaming into the night*

Not to mention that 2 + 2 = 5 for sufficiently large values of 2.

I had a math teacher prove that 1=0 once...

It involved dividing by (a-b) and then setting a=b. 😀