# Kinematics: Velocity and Speed

1. Feb 10, 2009

### Immy2000

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

A student travels at a constant velocity of 8.0 m/s north for 25 minutes, and then 5.0 m/s south for 15 minutes. Calculate the students...

a) Average Velocity (Answer 3.1 m/s North)
b) Average Speed (Answer 6.9 m/s)

2. Relevant equations

a) http://library.thinkquest.org/C0110840/Images/vel.gif

b) Vav = [Delta]d/[Delta]t

3. The attempt at a solution

a) Vav = [Delta]d/[Delta]t
[Delta]d = d2-d1
[Delta]t = t2-t1

t1= 0s
t2= 15 minutes = 900s

d1= 0m
d2= (5 m/s * 900s)

[Delta]d= 4500m - 0m = 4500m
[Delta]t= 900s - 0s

Vav = 4500 m
900 s
= 5.0 m/s [North]

????? I do not even understand velocity, so I just plugged in what I thought seemed logical as initial start I assumed was 0 and if you travel from 0 to x1 and then from x1 to x2, it would therefore be x2 being the final..thing and 0 being the initial..thing.

b) Vav = [delta]d\[Delta]t
[Delta]t = [t2 + t1]
[Delta]d = [d2*t2] + [d1*t2]
[Delta]d = (8 m/s * 1500 s) + (5 m/s * 900 s) = 16500 m
[Delta]t = (900 s + 1500 s)
Vav = 16500 m
2400 s
Vav = 6.875 = 6.9 m/s

?????? WOW I dont even know why it worked because I always though it was d2-d1 not d1+d2. I am so confused! Please, all help would be much appreciated. :)
I just cant seem to understand velocity which probably is the cause of my inability to differentiate what to do for either speed or velocity in general..
1. The problem statement, all variables and given/known data

Last edited by a moderator: Apr 24, 2017
2. Feb 10, 2009

### nothing19

Well for 1....
you want to take the distant traveled north. That's straightforward, just don't get stuck on units.
Remember, Distance = Speed * Time
Then measure the distanced traveled south, subtract that to have the total distance traveled.
You now will have a distance, and you know the time it took (25+15=40 min). D/T=V...so divide the distance by the time (convert back to seconds), and you'll get the right answer.

As far as average speed, that's even easier. You could just add the distance traveled in the first 25 to the distance in the other 15, and then divide that total distance by 40. Remember, speed doesn't take direction into account, just magnitude. Velocity is a vector and takes both magnitude and direction into account.

Hope that helps.

3. Feb 10, 2009

### Immy2000

Alright, I understand speed now, but if I read correctly, you firstly address velocity. Isn't velocity "change in displacement", not "change in distance" over "change in time"?

4. Feb 10, 2009

### nothing19

Yes, my bad. Should've said displacement, but the calculations are in fact calculating the displacement, not the distance.

5. Feb 10, 2009

### Immy2000

When I did the calculation I got 6.9 m/s which is not the answer to average velocity, but rather average speed.

6. Feb 10, 2009

### nothing19

Ah. you added the distances traveled. Remember when dealing with velocities you're dealing with vectors, and the vectors are in completely opposite directions here--the two must be subtracted.

7. Feb 10, 2009

### Immy2000

OH MY GOSH I LOVE YOU THANK YOU SO MUCH!!! Heh. No really, thanks, that really helped. Its so simple...I guess I just need to break it down and go from there. For now.