Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a constant speed of 30 mi /hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi /hr for the positions represented.

Incomplete question

Check attachment for the diagram of the problem.

Explanation:

The acceleration of the car A is given as

a=3ft/s²

Car B is rounding a curve of radius

r=440ft

Car B is moving at constant speed of Vb=30mi/hr.

Car A reach a speed of 45mi/hr

Note, 1 mile = 5280ft

And 1 hour= 3600s

Then

Va=45mi/hr=45×5280/3600

Va=66ft/s

Also,

Vb=30mi/hour=30×5280/3600

Vb=44ft/s

Now,

a. Let write the relative velocity of car B, relative to car A

Vb = Va + Vb/a

Then,

Using triangle rule, because vectors cannot be added automatically

Vb/a²= Vb²+Va²-2Va•VbCosθ

From the given graphical question the angle between Va and Vb is 60°.

Vb/a²=44²+66² - 2•44•66Cos60

Vb/a²=1936+ 4356 - 5808Cos60

Vb/a² = 3388

Vb/a = √3388

Vb/a = 58.21 ft/s

The direction is given as

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Va/SinA = Vb/SinB = (Vb/a)/SinC

66/SinA = 44/SinB = 58.21/Sin60

Then, to get B

44/SinB = 58.21/Sin60

44Sin60/58.21  = SinB

0.6546 = SinB

B=arcsin(0.6546)

B=40.89°

b. The acceleration of Car B due to Car A.

Let write the relative acceleration  of car B, relative to car A.

Let Aa be acceleration of car A

Ab be the acceleration of car B.

Ab = Aa + Ab/a

Given the acceleration of car A

Aa=3ft/s²

Then to get the acceleration of car B, using the tangential acceleration formular

a = v²/r

Ab = Vb²/r

Ab = 44²/440

Ab = 4.4ft/s²

Using cosine rule again as above

Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ

Ab/a²= 3²+4.4²- 2•3•4.4•Cos30

Ab/a²= 9+19.36 - 22.863

Ab/a² = 5.497

Ab/a = √5.497

Ab/a = 2.34ft/s²

To get the direction using Sine rule again, as done above

Using Sine Rule

a/SinA = b/SinB = c/SinC

i.e.

Aa/SinA = Ab/SinB = (Ab/a)/SinC

3/SinA = 4.4/SinB = 2.34/Sin30

Then, to get B

4.4/SinB = 2.34/Sin30

4.4Sin30/2.34 = SinB

0.9402 = SinB

B=arcsin(0.9402)

B=70.1°

Since B is obtuse, the other solution for Sine is given as

B= nπ - θ.   , when n=1

B=180-70.1

B=109.92°

answer and explanation;

composition refers to both the types of minerals within a rock and the overall chemical makeup of the rock while texture refers to the size and arrangement of the minerals or grains that make up a rock. both composition and texture can be used to classify rocks, such as igneous rocks. texture describes how the rock became solid or crystallized from liquid, melted rock. composition of a rock describes the minerals that make up a particular rock.

The independent variable in this experiment is the angle at which the light from the ray box strikes the mirror.hope this .