At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

* from the origin in quadrant 1 anticlockwise

How would I use components to find the avg. velocity @AllTehMaffs

Kidnapped? What a horrifying question.

Okay... I used 20° though

Does that matter?

So how would I use components?

Can you see how to get the lengths of the sides of that triangle?

Using sin and cos?

So to find the deltas I would...?

:) Not yet. you already have the points.
What's your change in x?

2.1sin20-(-1.33)

right, and you find your y the same way, then use Pythagoras to find C

I think I contradicted myself by saying we don't need angles, but you're good so far

This gives the displacement so I would have to divide by time?

yeah, and then that's your average velocity between those points

So then as far as acceleration goes?

That's where you need those other angles decided to overheat

wow, don't know how those messages got mashed up :P Sorry, my computer decided to overheat

|dw:1381700640583:dw|

Sorry 35 mph?

that's the angle. V1 was 35 mph

yeah

For v2 I would use 10sin20?

:/
\[a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }]\]

Not quite. Remember that when the person made the measurement, the car was going straight east.

So all of V2 is in the x direction

So what would it be?

What's v2x?

I'm not sure

How much of V2 is moving in the x direction?

2.1sin20?

|dw:1381701681185:dw|
|dw:1381701807535:dw|

Does that help?

I can't see the v2 stuff on the dawning

nooooo!! bummer. k
|dw:1381702127697:dw|

Does it make sense why we're not using the 20 degrees for this part of the calculation?

Yes

So 2.1cos0

2.1 miles is the distance away from the police station

??

2.1?

moving E at a speed of 10 mph"

V2 = ?

10

V2x - V1x = ?

Where did you get 1.33?

10-35cos66°

yes!! Now V2y - V1y = ?

0-35sin66

yes!!

now (V2x - V1x) / t = ax
(V2y - V1y)/t = ay
a^2 = (ax)^2 + (ay)^2

and a is your average acceleration

Should I use | |?

??

Abs value

for acceleration? It has to be positive anyway because you're squaring the term

Why is a squared?

When I divide by t it's 5/60 right?

ummm , yeah, I think so

5 min (1 hour/60 min)

yeah

I gotta run to work. This problem was mean!! :/

Ok so I get 387. What units should it be?

Ok!

the magnitude to the deceleration is (10-35)/5min

it might just be that, I don't know. Good mathin'!

When you have a chance just double check for me! :)

|dw:1381728262451:dw|