Average Speed & Velocity: Speed vs Velocity

[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.