Solving Velocity/Time Problems: Step-by-Step

[toyotadude]

Velocity/Time help? :(

Homework Statement

Hey guys, back again! This one's a real bugger on my online HW sheet :(:

1) The scale on the horizontal axis is 2 s per division and on the vertical axis 9 m/s per division. The particle has an initial position at 9 m.

[PLAIN]http://img20.imageshack.us/img20/3527/67419237.jpg

What is the position of the particle after the first 2 s? Answer in units of m.


2. The attempt at a solution

1) For the first problem, since it was asking for the first 2 seconds and the units are by 2's, I looked at the "1" on the x axis... and since it was a velocity time graph, it was a constant velocity ?

I guessed 36 (9 x 4, and looking at the 1...), and I got it wrong. But now I'm thinking about it, and since it is constant velocity it may be 72 (x2) because it lasted 2 seconds? (I'm probably getting it confused with the position/time graphs we're also doing ><)

 

[cepheid]

You're confusing velocity with position. The graph is a velocity-time graph. It tells you what the velocity was at any given time. Yes, the graph does indeed show that the velocity is constant throughout the motion (after all, the graph is flat). Therefore, you know that the particle's velocity was 35 m/s from t = 0 to t = 8 s. Having extracted this information from the graph, you no longer need to refer to it again.

Now comes the question: if the particle started at a position of 9 m, what was its position of the particle after 2 s? Well, this is a simple case of motion with a constant velocity. If a particle travels with a constant velocity of 36 m/s, how far does it travel in 2 s?

 

[toyotadude]

72 xD?

...which I just entered, and was denied?! :(

EDIT: OOOOH. 72 + 9 :O?

 

[cepheid]

Exactly. Remember that for motion at a constant velocity, the position, x, is given by:

x = x₀ + v*t

where x₀ is the initial position at time t = 0. This equation is not complicated. It is very easy to understand why it is that way. The equation is saying that:

final position = initial position + distance travelled

and

distance traveled = speed*time

Does this make sense to you? Anyway, in our case, we have:

x = 9 m + (36 m/s)*(2 s) = 9 m + 72 m

You should write out all of your calculations in full like this (including the units) from now on. It will really help you sort out what's going on.

 

[cepheid]

Oh yeah, and another way of solving the problem would have been to use the fact that the distance traveled over a certain time interval is the AREA underneath the velocity-time graph over that time interval. This area would have worked out to v*t = 72 m/s, same as before.

 

[toyotadude]

Makes a bit of sense, our teacher didn't go that indepth... but I somewhat get it :)

There were three parts to this question, and I've been slowly trying to figure them out (I've been absent a few days, and have no transportation for tutoring after school :\)

2) What is the velocity of the particle after the first 2 s. Answer in units of m/s?

My guess: Since it's a constant velocity of 36m/s... and it's constant... 36m/s? o.o

[Part] 3) What is the acceleration of the particle after the first 2 s? Answer in units of m/s^2.

My guess: There is no acceleration, because it's a constant velocity... 0?

 

[cepheid]

Makes a bit of sense, our teacher didn't go that indepth... but I somewhat get it :)

There were three parts to this question, and I've been slowly trying to figure them out (I've been absent a few days, and have no transportation for tutoring after school :\)

2) What is the velocity of the particle after the first 2 s. Answer in units of m/s?

My guess: Since it's a constant velocity of 36m/s... and it's constant... 36m/s? o.o

[Part] 3) What is the acceleration of the particle after the first 2 s? Answer in units of m/s^2.

My guess: There is no acceleration, because it's a constant velocity... 0?

You're correct on both counts! This stuff isn't hard. Answering these questions simply requires knowing what the definitions of these quantities are, and clearly you do.

EDIT: Or in this case, you can simply look at the graph. Since it's a velocity vs. time graph, it TELLS you what the velocity is at t = 2 s. So all you have to do is look at it.

For the acceleration, remember that the acceleration is the rate of change of the velocity time graph, which means it is equal to the slope of the graph. Here, the slope is zero. There is no rate of change of velocity.

So that's the graphical way to get the answers that were already intuitively obvious to you.

 

[toyotadude]

Yup, that's exactly how I figured it out! :)

Another brain teaser for you (and a complicated one for me... cause I don't know how to do it >_<)

[PLAIN]http://img696.imageshack.us/img696/9361/17430190.jpg

It is asking for: What is the average speed for the motion from point O to point B?

Biggest problem is, I don't know the formula for average speed. Assuming speed is m/s, then O to A is 2m/s, and A to B is (negative?) 1m/s... the average would be...?! It's just a guess, I'm not sure where to go from there.

It wouldn't be 1.5, as it's not listed on the answer choices... god, so confused >.<

EDIT: Part 2... What is the instantaneous velocity at point B?

My reasoning: I have honestly no idea, the teacher never even taught that (to my knowledge), and we're on the next unit already. Online lesson helps? :(

 

[toyotadude]

^ I thought it would be 0 for instantaneous velocity, being that B is... well at zero, and the velocity is 0/3... = 0, but I was still wrong?

And looking even further, it also asks for the instantaneous speed at point B. Speed is... velocity over time, correct? How would I calculate velocity for such a thing in the first place? (Velocity of the line, or the point..?!)

 

[toyotadude]

Bump? D: