Average Speed & Velocity: Speed vs Velocity

[UrbanXrisis]

I'm getting confused with speed and velocity. I know that speed does not tell direction and velocity does but here's the question:

A person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00m/s.

What is the average speed and average velocity?

I pretended that point A to B is 15m and calculated that the average speed would be 3.75s
Is there another way to calculate the speed if I had not made up a distance from A to B?

Also, what would the velocity be? Would it be the same as the speed? Since the velocity would be +5.00m/s for the first part then -3.00m/s how do I calculate the velocity?

 

[Amber]

As far as I know, velocity is speed with a direction.

I'm not sure on the rest though, sorry.

 

[HallsofIvy]

It is impossible to answer the first question without knowing the distance from A to B (or, same thing, the time required). Yes, if the distance was 15 m (good choice: its divisible by both 3 and 5: It took 15/5= 3 seconds to go from A to B and 15/3= 5 seconds to go from B to A. Total distance 30 m, time 8 sec so average speed is 30/8= 15/4= 3.75 m/s exactly what you said (except you had 3.75 s!).

On the other hand, if the distance from A to B were 30 m (still divisible by both 3 and 5!) the time to go from A to B would be 30/5= 6 s and the time to go from B to
A would be 30/3= 9 s. Now, he has gone 60m in 15 s. : average speed 60/15= 4 m/s.]

It is impossible to determine the average speed without knowing the distance.

Answering the second question is easy! The person's DISPLACEMENT (distance from initial point) is 0 because he returned to his initial point. No matter what the time, t, required his average VELOCITY is 0/t= 0.

 

[OsiriS^]

As far as I know, velocity is speed with a direction.

I'm not sure on the rest though, sorry.

Yeah, velocity is a vector so has magnitude and direction where as speed is a scalar and has magnitude only.

Speed = Distance/Time

Velocity = Displacement/Time

 

[UrbanXrisis]

what is the difference between the average velocity and velocity?

If the person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to a point before point A at a constant speed of 3.00m/s, what would the difference between average velocity and velocity be?

 

[Pyrrhus]

Halls, i think he can do it.

Ok so let's do this:

info
He covered a distance x

V_{1} = 5 m/s

V_{2} = 3 m/s

Using this equation

V = \frac{d}{t}

t_{2} = \frac{x}{V_{2}}

t_{1} = \frac{x}{V_{1}}

Remember Average Speed is Total Distance divided by Total Time so

t_{1} + t_{2} = T_{total}

2x = D_{total}

so

V_{ave} = \frac{2x}{\frac{x}{V_{2}} + \frac{x}{V_{1}}}

solving it

V_{ave} = 3.75 m/s

 

[Pyrrhus]

what is the difference between the average velocity and velocity?

If the person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to a point before point A at a constant speed of 3.00m/s, what would the difference between average velocity and velocity be?

Velocity or Instant Velocity, it's just dv/dt, which means a instant value, which could be different before the body reached it or after it reached it, it's like a velocity at x point.

Average velocity on the other hand it's displacement divided by time change of the whole movement, from where it started to where it finished. In your case the body started a 0 so the displacement vector went from that and then returned to the exact point, so the displacement became 0.

Velocity is a vector while Speed is a Scalar.

 

[UrbanXrisis]

What if I started 5 meters away from the origin and came towards the origin at 5m/s for 2 seconds. Then what would the velocity be? would it be -5m/s?

However, my displacement is 5m so would it be -2.5m/s?

 

[Pyrrhus]

What if I started 5 meters away from the origin and came towards the origin at 5m/s for 2 seconds. Then what would the velocity be? would it be -5m/s?

However, my displacement is 5m so would it be -2.5m/s?

Your Average Velocity will be 0, still. Your Displacement was 0.

You covered a 5 meter a amount, and the you covered the same.

it will be like 0-0 =0

Remember

\Delta x = x - x_{o}

So you initial X will be 0 and your final X will be 0 too!

 

[Pyrrhus]

What if I started 5 meters away from the origin and came towards the origin at 5m/s for 2 seconds. Then what would the velocity be? would it be -5m/s?

However, my displacement is 5m so would it be -2.5m/s?

Remember displacement only sees final and initial, if you came from the origin at any speed for any time and came back to the origin again, your average velocity will still be 0!

 

[Pyrrhus]

What if I started 5 meters away from the origin and came towards the origin at 5m/s for 2 seconds. Then what would the velocity be? would it be -5m/s?

However, my displacement is 5m so would it be -2.5m/s?

Oh i misread , yes. (to the -2.5 m/s)

 

[UrbanXrisis]

I understand that but re-read the question, I came back towards the origin for 2 seconds at 5m/s, that means I have gone past the origin which is only 5m away. Therefore, I have a displacement of -5m. So, what would the velocity be? would it be -5m/s?

Since my displacement is -5m, would my velocity ve -2.5m/s?

 

[Pyrrhus]

I understand that but re-read the question, I came back towards the origin for 2 seconds at 5m/s, that means I have gone past the origin which is only 5m away. Therefore, I have a displacement of -5m. So, what would the velocity be? would it be -5m/s?

Since my displacement is -5m, would my velocity ve -2.5m/s?

read above i noticed later

 

[UrbanXrisis]

Okay, still confused :P

Walking away from the origin gives a positive velocity. Walking towards the origin gives a negative velocity. What if I was behind the origin, and walking towards it? is it still positive? What if I was behind the origin and walked away? That's walking away from the origin, so that means it's positive? so would the velocity to my lasy question be positive 2.5m/s?

 

[Pyrrhus]

Okay, still confused :P

Walking away from the origin gives a positive velocity. Walking towards the origin gives a negative velocity. What if I was behind the origin, and walking towards it? is it still positive? What if I was behind the origin and walked away? That's walking away from the origin, so that means it's positive? so would the velocity to my lasy question be positive 2.5m/s?

It's just a sign convention you pick.. The standard is right and up positive and down and left negative.

If you walk towards the origin the vector is pointing left, so it's negative under that convention. It will be positive under that convention wallking toward the origin from behind (i assume you mean down), it will be negative because of that convention walking away from the origin if you were behind. It will be negative.

 

[UrbanXrisis]

What if I walked away from the origin 10m to the right for 2s, then walked back 5m in 3 seconds. My total displacement is 5m, my time is 5s so my velocity is 1m/s?

 

[Pyrrhus]

Remember is displacement divided by time change, so only 1 second.

 

[UrbanXrisis]

but it took me 5s to get there, how is it 1s?

 

[Pyrrhus]

but it took me 5s to get there, how is it 1s?

Well by applying the definition of Average Velocity.

Thinking it logically, you walked for 3 seconds to a x place then in the return it took you 2 seconds, so logically you might think 3+2=5, will be total seconds taken, but Average Velocity actually means a slope of a triangle with opposite side displacement and a adjacent side of change of t. The problem i see with you, is you are interpreting velocity the same as speed, in physics they do not mean the same!.

 

[UrbanXrisis]

OKay, I'm just making up random questions so I can understand this concept.

If I started 10m away from the origin, I walked 5m towards it in 1s. Would my velocity be negative 5m/s or positive?

Lets say I started 10m away from the origin, and walked to it in 2s. My velocity is zero?? Would it be -5m/s since I STARTED away from the origin?

I tested this on a motion sensor...I walked towards it at a constant speed and it showed a negative velocity. However, if I'm returning to the origin, wouldn't the velocity be zero?

 

[Pyrrhus]

OKay, I'm just making up random questions so I can understand this concept.

If I started 10m away from the origin, I walked 5m towards it in 1s. Would my velocity be negative 5m/s or positive?

Lets say I started 10m away from the origin, and walked to it in 2s. My velocity is zero?? Would it be -5m/s since I STARTED away from the origin?

I tested this on a motion sensor...I walked towards it at a constant speed and it showed a negative velocity. However, if I'm returning to the origin, wouldn't the velocity be zero?

1) -5m/s

2) -5 m/s

3)No, because there's a negative displacement.

Velocity is only 0 when you walk from a x point and then return to it.

 

[UrbanXrisis]

what if I began from the origin, walked 10m in 2s, then walked 5m back in 2s...

v=5m/2s-2s? change in time ?

 

[Pyrrhus]

what if I began from the origin, walked 10m in 2s, then walked 5m back in 2s...

v=5m/2s-2s? change in time ?

If you began from the origin it will have 0s at the origin.

and i know what you're trying to do displace divided by 0, but that will be an impossible case, at least in Classical Physics, because that will give infinity.

 

[UrbanXrisis]

Let me restate...I first walk 10m forward, which takes me 2 sec, then at 2.5m/s, I walk 5m towards the origin (which will take me 2s).

I am still 5m away from the origin

what is my velocity?

 

[Pyrrhus]

Let me restate...I first walk 10m forward, which takes me 2 sec, then at 2.5m/s, I walk 5m towards the origin (which will take me 2s).

I am still 5m away from the origin

what is my velocity?

read above.

 

[UrbanXrisis]

but wouldn't it be

v=5m/4s?

displacement from the origin is 5m, time it took is 4s

 

[Pyrrhus]

but wouldn't it be

v=5m/4s?

displacement from the origin is 5m, time it took is 4s

It will be the third and first case, to use with the definition, you need to have 0s somewhere an origin.

 

[Pyrrhus]

Well I'm going to have lunch, go do some problems and see if you can understand the concept that way, the best way is doing problems.

 

[UrbanXrisis]

What if I walked away from the origin 10m to the right for 2s, then walked back 5m/s in 3 seconds. My total displacement is -5m, my time is 1s so my velocity is 5m/s?

 

[UrbanXrisis]

alright, thanks for the help!

 

[swingkids]

If you're going for 20 minutes at 20 miles/hr in the direction of 15° east of north, and then going for 35 minutes at 25 miles/hr in the direction of 20° south of west, what's the average velocity and average speed after 55 minutes?

 

[HallsofIvy]

Well by applying the definition of Average Velocity.

Thinking it logically, you walked for 3 seconds to a x place then in the return it took you 2 seconds, so logically you might think 3+2=5, will be total seconds taken, but Average Velocity actually means a slope of a triangle with opposite side displacement and a adjacent side of change of t. The problem i see with you, is you are interpreting velocity the same as speed, in physics they do not mean the same!.

No, this is completely wrong. "Average Velocity" is the (vector) displacement divided by the total time (time is not a vector).

If you walk 10 m forward in 3 seconds, then walk 5 meters back in 2 seconds, you are now 10-5= 5 meters in front of your original position an you have taken a total of 5 seconds to do that: you "Average Velocity" is 5/5= 1 m/s.

If you had asked "Average Speed" then you would calculate that you walked a total of 15 m (10 out and then 5 back) in 5 seconds so that your average speed is 15/5= 3 m/s.

 

[HallsofIvy]

If you're going for 20 minutes at 20 miles/hr in the direction of 15° east of north, and then going for 35 minutes at 25 miles/hr in the direction of 20° south of west, what's the average velocity and average speed after 55 minutes?

First it is not a good idea to add your own question to someone else's thread- start your own thread.

20 minutes is 1/3 of an hour so "20 minutes at 20 miles/hr" is 20/3 miles/
35 minutes is 35/60= 7/12 of an hour so "35 minutes at 25 mile/hr" is (7/12)(25)= 175/12 miles.

For the average velocity, start by drawing a picture: Taking north "up"
and east "right", draw a line at 15° to the "right of up" and of length
20/3. From there draw a line 20° "below a horizontal line running left" of length 175/12 and connect the endpoints with a third line (the unknown displacement vector) forming a triangle.

Now there are two ways to handle this.

1) It should be easy to see that the angle of the triangle opposite the unknown side has size 90-20-15= 55°. Use the cosine law to find the length of the third, unknown, side. That is the magnitude of the displacement vector. Dividing by the total time: 20+35= 55 minutes= 55/60= 11/12 hour. That is the magnitude of the average velocity vector. You can now use the sine law to find the angle that third side makes with either "north" or "west" to complete the vector.

2) I would be inclined to do this by "components". Set positive x "east" and positive y "north". A vector of length 20/3 making an angle of "15° east of north" has x-component (20/3)cos(15°) and y-component (20/3)sin(15°). A vector of length 175/12 and making an angle of "20° south of west" has x-component -(175/12)cos(20°) and y-component -(175/12)sin(20°).
Now add components to find the components of the displacement vector. Dividing both components of that by the total time, 11/12 hour, to get the average velocity vector.

"Average speed" is easy: we already know we went a total of 20/3+ 175/12= (80+ 175)/3= 255/3 miles in 11/12 hour. Divide the total distance by the total time to find average speed.

 

[Pyrrhus]

weird, my physics book has it as delta X/ delta T, of course the depends if you're calculating the average velocity of the whole movement..., but if so thanks for pointing it out.

 

[swingkids]

"First it is not a good idea to add your own question to someone else's thread- start your own thread."

I did. No one replied.

Thanks, though I had to do it yesterday.