Clayton State University
Department of Natural Sciences
September 8, 2005
Physics 1111 – Quiz 2
Name __SOLUTION___________________________________
1. A new car manufacturer advertises that their car can go "from zero to sixty in 8 s".
This is a description of
a. Average speed.
b. Instantaneous speed.
c. Average acceleration.
d. Instantaneous acceleration.
2. Suppose that an object is moving with a constant velocity. Make a statement
concerning its acceleration.
a. The acceleration must be constantly increasing.
b. The acceleration must be constantly decreasing.
c. The acceleration must be a constant non-zero value.
d. The acceleration must be equal to zero.
3. During the first 8 seconds
a. C has decreasing velocity, D has increasing velocity.
b. C and D both have decreasing velocities.
c. C and D have the same velocity.
d. C has the same average velocity as D.
Department of Natural Sciences
Clayton State University
September 5, 2007
Physics 1111 – Quiz 2
Name ______SOLUTION_______________________________
1. If you start from the Art Gallery, travel to the Café, and then to the Bakery, what is your
displacement?
a. 6.50 km.
b. 2.50 km.
c. 10.5 km.
d. -1.50 km.
2. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour,
what is your average speed?
a. 2.50 km/hr.
b. 4.00 km/hr.
c. 9.00 km/hour.
d. 10.5 km/hr.
3. Given the position-versus-time graph for a basket ball player traveling up and down the
courting a straight-line path find the instantaneous velocity of the player
a. At t = 6.00 s.
V = 0 (tangent is horizontal at t = 6.00 s)
b. At t = 9.00 s
V = rise/run = (-6.00 m)/(4.00 s) = -1.50 m/s


Department of Natural Sciences
Clayton State University
January 24, 2007
Physics 1111 – Quiz 2
Name ____SOLUTION_________________________________
1. If you start from the Bakery, travel to the Café, and then to the Art Gallery, what is your
displacement?
a. 6.50 km.
b. -2.50 km.
c. 10.5 km.
d. -1.50 km.
2. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour,
what is your average speed?
a. 6.50 km/hr.
b. 2.50 km/hr.
c. 9.00 km/hour.
d. 10.5 km/hr.
3. Given the position-versus-time graph for a basket ball player traveling up and down the
courting a straight-line path find the instantaneous velocity of the player at t = 4.00 s.


V = (6.00 m - 0 m)/(4.00 s - 0 s) = 1.50 m/s

Department of Natural Sciences
Clayton State University
January 23, 2008
Physics 1111 – Quiz 2
Name ___SOLUTION__________________________________
1. If you start from the Bakery, travel to the Café, and then to the Art Gallery, what is your
displacement?
a. 6.50 km.
b. 2.50 km.
c. -4.00 km.
d. -2.50 km.
2. If you start from the Cafe, travel to the Bakery, and then to the Art Gallery, in 1.00 hour,
what is your average velocity?
a. 2.50 km/hour.
b. 6.50 km/hour.
c. - 6.50 km/hour.
d. - 10.5 km/hour.
3. Given the position-versus-time graph for a basketball player traveling up and down the
courting a straight-line path find the instantaneous velocity of the player
a. At t = 2.00 s,
V = rise/run = (6.00 m)/(4.00 s) = 1.50 m/s
b. At t = 5.00 s,
V = rise/run = (-3.00 m)/(2.00 s) = -1.50 m/s
c. At t = 8.00 s.
V = 0 (tangent is horizontal at t = 8.00 s)


Department of Natural Sciences
Clayton College & State University
May 25, 2004
Physics 1111 – Quiz 2
Name _____SOLUTION________________________________
1. A bird, accelerating from rest at a constant rate, experiences a displacement of 28.0 m in 11.0
seconds. What is its average velocity?
a. 1.73 m/s.
b. 2.55 m/s.
Vav = x/t = (28.0 m)/(11.0 s)
c. 3.41 m/s.
d. zero.
4. 2. Given that a = 2.00 m and b
= 4.00 m, find:
a. The length of side c.
c2 = a2 + b2
c2 = (2.00 m)2 + (4.00 m)2 = 20.0 m2
c = 4.47 m
b. Cosine of angle 

cos(b/c = (4.00 m)/(4.47 m) = 0.895


c. Tangent of angle 

tan (a/b = (2.00 m)/(4.00 m) = 0.500
d. Angle .
 cos-1( 26.5o

Department of Natural Sciences
Clayton College & State University
June 8, 2005
Physics 1111 – Quiz 2
Name __SOLUTION___________________________________
1. A vehicle accelerates from 0 to 30.0 m/s while undergoing a straight line displacement of
45.0 m. What is the vehicle’s acceleration if its value may be assumed constant?
a. 2.0 m/s2.
V2 = V02 + 2 a (x – x0)
2
x0)] = [(30.0 m/s) – 0] / [2 x 45.0 m] = 10.0 m/s2
b. 5.0 m/s2.
a = (V2 - V02 )/[2(x –
c. 10.0 m/s2.
d. 15.0 m/s2.
2. Which of the following is equivalent to acceleration?
a. Displacement.
b. Rate of change of displacement.
c. Velocity.
d. Rate of change of velocity.
3. A bird acceleration from rest at a constant rate experiences a displacement of 28.0 m in
11.0 s. What is its average velocity?
a. 1.70 m/s.
Vav = x / t = (28.0 m)/ (11.0 s)
b. 2.55 m/s
c. 3.40 m/s
d. 0 m/s.
Department of Natural Sciences
Clayton State University
June 6, 2007
Physics 1111 – Quiz 2
Name ______SOLUTION_______________________________
1. If you start from the Bakery, travel to the Café, and then to the Art Gallery, what distance
did you travel?
a. 6.50 km.
b. -2.50 km.
c. 10.5 km.
d. -1.50 km.
2. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour,
what is your average velocity?
a. 2.50 km/hr.
b. 4.00 km/hr.
c. 9.00 km/hour.
d. 10.5 km/hr.
3. Given the position-versus-time graph for a basket ball player traveling up and down the
courting a straight-line path find the instantaneous velocity of the player
a. At t = 2.00 s.
V = (6.00 m)/(4.00s) = 1.50 m/s
b. At t = 8.00 s
V=0


Department of Natural Sciences
Clayton State University
January 28, 2009
Physics 1111 – Quiz 2
Name ____SOLUTION_________________________________
1. If you start from the Bakery, travel to the Art Gallery, and then to the Cafe, what is your
displacement?
a. 6.50 km.
b. 4.00 km.
c. 10.5 km.
d. -1.50 km.
2. If you start from the Bakery, travel to the Art Gallery, and then to the Café, in 1.00 hour,
what is your average velocity?
a. 4.00 km/hr.
b. 6.50 km/hr.
c. -9.00 km/hr.
d. 10.5 km/hr.
3. Given that a = 2.50 m and b = 4.50 m, find:
a. The length of side c.
a2 + b2 = c2
c = (a2 + b2)1/2
c = ((2.50 m)2 + (4.50 m)2)1/2 = 5.15 m
b. The sine of angle 

sin  = a/c = (2.50 m)/(5.15 m) = 0.485
c. The tangent of angle 

tan  = a/b = (2.50 m)/(4.50 m) = 0.556

d. Angle 


 = tan-1 (0.556) = 29.1o


Clayton State University
Department of Natural Sciences
January 27, 2010
Physics 1111 – Quiz 2
Name _SOLUTION____________________________________
1. A car moving due north along a straight highway changes its speed from 20.0 m/s to
15.0 m/s. The car’s acceleration is directed
a. North.
b. East.
c. South.
d. West.
2. Suppose that an object is moving with a constant velocity. Make a statement
concerning its acceleration.
a. The acceleration must be constantly increasing.
b. The acceleration must be constantly decreasing.
c. The acceleration must be a constant non-zero value.
d. The acceleration must be equal to zero.
3. During the first 8 seconds
a. A moves to the right, B moves to the left.
b. C and B both have negative velocities.
c. C moves to the left, A moves to the right.
d. A and B are at rest.






