physics question

[renegade05]

A car travels three quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. The initial velocity is west and the final velocity is south.
a) Find its average velocity for this trip.
b) What is the cars average acceleration during these 3.0 s?
c) Explain how a car moving at constant speed has a nonzero average

Apparently the the answer to a is 9.4 m/s 45 degrees north of east. B is 15m/s2 SE. C is becasue a changing the direction of the velocity requires an acceleration.

How did they come up with these answers? I am trying to figure it out but cant. Wouldnt ave vel be the circumfirunce of the circle multipilied by the fraction of a rev. ((40pi)(3))/(4) and then divided by the 3.0 sec to find m/s?
I get 31m/s

Please help with these. Thanks.

 

[tkhunny]

Original Velocity is West. Two choices
-----1) Start at the bottom and head clockwise
-----2) Start at the tom and head anti-clockwise
Ending 3/4 of the way around
-----1) Start at the bottom and head clockwise and end on the right headed south
-----2) Start at the tom and head anti-clockwise and end on the right headed north
This rules out choice #2

Total Displacement from (0,-20) to (20,0) is \(\displaystyle \sqrt{20^{2}+20^{2}}\;=\;20\cdot\sqrt{2}\)

Average Displacement, then: \(\displaystyle \frac{20}{3}\cdot\sqrt{2}\;=\;9.428\)

Travelling from (0,-20) to (20,0) is pretty obviously 45º East of North

Now what?

 

[renegade05]

Thats the part I'm really stuck on. I have no idea how the answer is 15m/s for acceleration.

 

[tkhunny]

This is a wonderful problem to force you to consider the difference between speed and velocity. The answer to part c) is awesome. "Speed is not velocity, Silly!" I'd write it on an exam. :shock:

Distance = Rate * Time

Run this through your mind.

Average Velocity = (Total Displacement) / Time = (Change in Position) / Time

This works for acceleration, too!

Average Acceleration = (Change in Velocity) / Time

3/4 trip around the circle is \(\displaystyle \frac{3}{4}\cdot 2\cdot\pi\cdot 20 = 30\cdot\pi\)

In 3 seconds, what is the constant speed on the circle?

If you know where it starts and where it stops, you are very nearly done.

 

[renegade05]

Well the constant speed would be 30pi / 3 seconds. right? 10 pi...

now what...

 

[tkhunny]

See that big square root with the squared 20s? Do something like that.

 

[renegade05]

Well i know the answer and even the procedure to do...

It shows \(\displaystyle \sqrt{1^2+(-1)^2}\)
Where the heck did these 1's come from?

It also shows to square the time.

I thought Aave was just change in velocity over change in time.

....

 

[tkhunny]

Hopefully, there was a 10pi in front of that radical with the '1's.