1. Feb 2, 2006

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I need help on these problems. Thank You.

1. A particle moves according to the equation
x = (10 m=s*s) t*t where x is in meters and t is
in seconds.

a. Find the average velocity for the time in-
terval from t1 = 2:38 s to t2 = 3:36 s. Answer
in units of m=s.

b.Find the average velocity for the time interval
from t1 = 2:38 s to t3 = 2:48 s. Answer in
units of m=s.

2. Consider the acceleration of gravity to be
10 m/s*s :

a. What is the magnitude of the instantaneous
velocity (speed) of a freely falling object 38 s
after it is released from a position of rest?

b. What is its average speed during this 38 s
interval? Answer in units of m/s.

c. How far will it fall during this time? Answer
in units of m.

2. Feb 2, 2006

Staff: Mentor

What do you mean by "m=s"? Do you mean m/s?

3. Feb 2, 2006

Hurkyl

Staff Emeritus
What have you done on these problems? Even if you've gotten barely anywhere, it's still better than nothing. (It's homework help, not homework answerbook)

4. Feb 2, 2006

yes i mean m/s.

5. Feb 2, 2006

I just need to know how to start them not the answers.

6. Feb 2, 2006

Hurkyl

Staff Emeritus
Surely you've had an idea? Thought about equations that might be useful? Thought about similar problems you could solve? Thought about what you can figure out, even if it's not what the problem is asking for? Thought about what information would let you answer the problem, even if you don't know how to get that information? Or even just substituted in the definition of a term?

Many of these are applicable to nearly every problem you will face -- so if you get in the habit of thinking about them, you will always be able to start a problem!

Last edited: Feb 2, 2006
7. Feb 2, 2006

Staff: Mentor

Yeah, like what is the difference between "average" and "instantaneous" velocity? What basic mathematical technique do you use to calculate an average?

8. Feb 3, 2006

sporkstorms

Another hint, in case this part is giving you trouble:

When they say the magnitude, they mean the overall positive value in whatever direction the velocity is.

For example, if we're working in two dimensions and something has an x-velocity of 2, and a y-velocity of 3, the velocity is pointing diagonally "up" in an x-y graph. The magnitude of the velocity would be the length of the line you graph from the origin to the point (2,3).
22 + 32 = v2
v = sqrt(13).

What about in one dimension, as with your problem? Well, since there's only a velocity component in one direction, the magnitude is the absolute value of that component. So if you have only a y-velocity, as you will in this problem, it will be negative since the object is falling. But the magnitude will be positive, because there is no such thing as a negative length.

9. Feb 3, 2006

vaishakh

Average velocity deals with time intervals while instantaneous velocity deals with time instants.
Like average velocity = displacement in the time interval/time interval
instantaneous velocity = velocity at that instant, orthe velocity at which the particle would continue to move if the particle acceleration is put to 0 at that point.
I cannot do further help when you are not doing just by not posting the copy of your arguement of the solutions and when one of the mentors are refusin to help you? I cannot help you neglecting Hurkyl advice.

10. Feb 3, 2006

Hurkyl

Staff Emeritus
I'm not refusing to help -- my goal is to respond to this problem in a way that helps the OP solve the next problem! (IMHO that's what this forum is all about) So, I'm trying to instill into the OP the notion that he should be trying these basic solving strategies when he doesn't know what to do.

If he does some of these basic strategies, and is still stuck, I would help him along further!