Your Name:  _______________________                              PHY203

                                                                                                            Final Exam

                                                                                                            Wed., 12/14/05

 

 

 

Lecture Time:         9 a.m             1p.m.             2p.m.       3p.m.       Honors         

 

 

Part 1

 

Part 1                                                            

 

 

1-10    ________     (out of 50)                

(5 x ____ )                                                   

 

 

11       ________     (out of 50)              

 

 

Total    _______                                         

 

 

Multiple choice answer sheet-shade in correct answers below

(one choice per problem):

 

 

 

1

 

2

 

3

 

4

 

5

 

6

 

7

 

8

 

9

 

10

 

a

 

 

 

 

 

 

 

 

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

c

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

 

 

 

e

 

 

 

 

 

 

 

 

 

 

 

For questions 1 and 2:

At t = 0, the speed of an object starting from the origin is 40 m/s at an angle of 45o with respect to the x-axis.  At t=5 sec., the particle is at x= -15m and y= 20m with a speed of 35 m/s at an angle of 30o with respect to the x-axis.

 

1.         Find the average velocity over the time interval:

a.         -1 m/s

b.        15 m/s

c.         37.5 m/s

d.         -3 m/s i +4 m/s j

e.         3 m/s i –4 m/s j

 

 

2.         Find the average acceleration over the time interval:

a.         -1 m/s2

b.        15 m/s2

c.         -15 m/s2

d.         5 m/s2 i –10.8 m/s2 j

e.         1 m/s2 i –2.2 m/s2 j

 

For questions 3-5:

A man runs 5 km in 30 min., then walks halfway back to the starting point in an additional 30 min.

 

3.         Find the total displacement:

a.         2500 m

b.        -2500 m

c.         7500 m

d.         -7500 m

e.         0 m

 

4.         Find the total average speed:

a.         0 m/s

b.        0.69 m/s

c.         -0.69 m/s

d.         2.1 m/s

e.         -2.1 m/s

 

5.         Find the total average velocity:

a.         0 m/s

b.        0.69 m/s

c.         -0.69 m/s

d.         2.1 m/s

e.         -2.1 m/s

 

Given the choices above, answer the following questions, 6-10:

 

6.         Given a plot of x vs. t as shown below, which of the graphs above depicts a plot of v vs. t:            a., b., c., d., or e. none of the above?

7.         Given the same plot of x vs. t as shown in #6, which of the graphs above depicts a plot of a vs. t:        a., b., c., d., or e. none of the above?

 

 

8.   Given a plot of v vs. t as shown below, which of the graphs above depicts a plot of a vs. t:       a., b., c., d., or e. none of the above?

9.         Given a plot of x vs. t as shown below, which of the graphs above depicts a plot of v vs. t:            a., b., c., d., or e. none of the above?

10.      Given a plot of a vs. t as shown below, which of the graphs above depicts a plot of v vs. t:            a., b., c., d., or e. none of the above?


Your Name:  _______________________

11.  A projectile is shot in a direction 35 degrees above the horizontal at an initial

speed of 50m/s, as shown above. (Note: Use 9.81 m/s2 for the

magnitude of g, the acceleration due to gravity.)

a.  Write the initial velocity vector, vo, in vector notation.

 

 

 

 

 

 

 

 

b.  Calculate the time in sec. that it will take for the projectile to reach its highest point.

 

 

 

 

 

 

 

 

c.  Calculate the height of the projectile at its highest point

 

 

 

 

 

 

 

d.  Calculate the x position of the projectile at its highest point

 

 

 

 

 

 

 

 

e.  Write down the velocity of the projectile at its highest point in correct vector

notation.

 

 

 

PHY203

Exam #1

Crib Sheet

Chapters 1-3

 

 

speedAV = (total distance traveled)/Dt

 

Dx = x2 – x1 (displacement)

 

vAV = Dx/Dt

 

aAV = Dv/Dt

 

vInst = dx/dt

 

aInst = dv/dt

 

 

For constant acceleration:

 

xf = xo + vot + (1/2)at2

 

vf = vo + at

 

Vf2 = vo2 + 2a(xf - xo)

 

 

g = 9.81 m/s2, the acceleration due to gravity

 

 

 

For a vector A with magnitude A and direction q (measured counterclockwise with respect to the x-axis):

 

Ax = Acos(q)

 

Ay = Asin(q)

 

A = (Ax2 + Ay2)1/2

 

tan(q) = Ay/Ax

 

quadratic eq. sol.:      If ax2 + bx + c = 0; then x = -b+(b2-4ac)1/2/(2a)