# Need help on Velocity

1. Aug 31, 2010

### mogibb1

This is my first physics class (and first science class in almost 20 years). This maybe a stupid question, but I don't really know if the book is looking for just a formula or what.

1. The problem statement, all variables and given/known data

A ball is thrown straight up with enough speed so that it is in the air for several seconds.
(a) What is the velocity of the ball when it reaches its highest point
(b) What is its velocity 1 s before it reaches its highest point?
(c) What is the change in its velocity during this 1-s interval?

2. Relevant equations

3. The attempt at a solution

I am assuming that for (a):
Velocity = 0 because it has reached its highest point

For (b) I am assuming that:

V = -1 because the ball is going straight up

For (c) I'm completely lost

As I said, I'm not real sure about this and the professor did not really explain this stuff.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 31, 2010

### CompuChip

The acceleration due to earth gravity is usually denoted g, its value is about 9.8 m/s2.
Can you explain what acceleration has to do with velocity and what the unit (m/s/s) means?

3. Aug 31, 2010

### mogibb1

The acceleration due to earth gravity is usually denoted g, its value is about 9.8 m/s2.
Can you explain what acceleration has to do with velocity and what the unit (m/s/s) means?

I know that velocity is speed and direction. I think that the (m/s/s) is meters per second squared....which is meters per second being travelled (up or down) and the other is the direction travelled per second.

4. Aug 31, 2010

### CompuChip

Hmm, sounds like you may not have had this yet (although I have no idea how you should answer the question then)... but it is important so I'm still going to explain this.

The thing that is important in physics is the position of an object. For example, a car is at a latitude/longitude of 30 degrees north and 41 degrees 12 minutes west, or a block has slid 3 cm down the 30°/1 meter ramp or a ball is at the point (-1, 3, 0) in our (x, y, z)-coordinate system. The latter is the most typical description in physics: choose an x- and y-axis and possibly z-axis (like when you are drawing a graph) and describe positions in two or three coordinates.

Of course, physics is not interesting if everything in our world would sit there forever, so we are going to look at movement. In other words, we need to describe how these positions change.

Velocity is the thing that describes how fast the position of an object changes. For simplicity, think of one dimension first. Suppose that something (a car, a block, a ball) moves 5 meters in 2 seconds at a constant velocity. Then the speed in this case would be 5 meters / 2 seconds = 5/2 meters / second = 2.5 m/s (I am abbreviating meters and seconds now). So in 4 seconds, it would move (2.5 m/s) x (4 s) = 10 m, in one minute it would move (2.5 m/s) x (60 s) = 150 m. Note that by "it would move" I actually mean: "the position would change by ...".

Of course, in the real world, velocities are hardly ever constant. Suppose that you are in a car which has stopped for a red light, and once the light turns green it smoothly increases its velocity to 25 m/s (that is 90 km/h or about 55 mph). Of course, in everyday life, you would say it accelerates from 0 to 90. Physicists also use this word: the car undergoes acceleration (and by "smoothly" I mean that the acceleration is constant). What is this acceleration? Well, basically, it is just the change of velocity, just as velocity itself was the change of position. So if the velocity changes by 25 m/s in, say, 5 seconds then the acceleration of the car will be (25 m/s) / (5 s) = 25/5 (m/s)/s = 5 m/s².
It is 5 meters per second-squared, meaning that every second, the velocity changes by 5 meters per second.

Now on earth, if you drop something off a table or a cliff, it accelerates at a constant acceleration. The precise magnitude of this acceleration depends on where you are, in my country it is about 9.812 m/s² [on the equator it is 9,780 m/s², on the poles about 9,832 m/s² and on the moon it is only 1,63 m/s² - about one sixth of the earth value].
That means that if I drop a ball, after a second it will have a velocity of 9.812 m/s. Ok, let's not make it complicated and pretend that g = 10 m/s², so it will have a velocity of 10 m/s. In the second second, the velocity will increase by 10 m/s, so it will be 20 m/s. After 2.5 seconds, the velocity will be 25 m/s (it has increased by 5 m/s in the next half second).

In the beginning I talked a bit about coordinates... what I forgot to mention there was that coordinates can be both positive and negative. You have a lot of freedom there... you can say "I will call upwards positive". Then v = 10 m/s means that the velocity (or speed in that direction) is 10 m/s upwards, and -10 m/s means that it goes down at 10 m/s. In that case, technically, you should say that the acceleration due to earths' gravity is -10 m/s² (or 9.81 m/s² whatever the appropriate number at your position is), because of course every second the velocity changes in the downward direction. This means that if you shoot something up at 15 meters per second (velocity 10 m/s) then after one second the velocity will be 15 m/s - 10 m/s = 5 m/s: the object (bullet, ball) is still going up, but slower. One second later, the velocity is 5 m/s - 10 m/s (or 15 m/s - 2 x 10 m/s) = -5 m/s: the object is now moving in the other direction, it is falling down. Of course, in between there is a point somewhere, where the velocity turns from positive to negative. It is precisely zero there (it just "hangs" there, in a manner of speaking) and thi s is, as you correctly thought, the highest point (it stopped moving up, and is about to start moving down).

5. Aug 31, 2010

### mogibb1

Ok. I think I may have it now. I appreciate the help. :)

This is my first week in this class and am a little overwhelmed but I shall persevere. Thanks again.