No Title
PHYSICS I
Solution to Exam 1
                                                                                                           
Thursday 9/17/15 Name
O101: periods 1,2
JFW CM:
Section:














                    
                                         (1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
  1. In the items below, circle the correct answers.

      (a) [5 pts] What is the value of p(8.104)2 written to the correct number of significant figures?
      A) 206.324
      B) 206.323
      C) 206.3
      D) 206
      E) 200
      ANS: C).
      (b) [5 pts] What is 56+32.00/(1.2456+3.45) written with the correct number of significant figures?
      A) 62.8
      B) 62.812
      C) 62.81
      D) 63
      E) 62.8123846
      ANS: D).
      (c) [5 pts] The height of the ceiling in a typical home, apartment, or dorm room is closest to
      A) 100 cm
      B) 200 cm
      C) 400 cm
      D) 500 cm
      Assuming an 8 ft ceiling, the answer is B).
      (d) [10 pts] A rabbit trying to escape a fox runs north for 8.0 m, darts northwest for 1.0 m, then drops 1.0 m down a hole into its burrow. What is the magnitude of the net displacement of the rabbit?
      A) 8.8 m
      B) 8.1 m
      C) 66 m
      D) 10 m
      ANS: A).

    2. (a) [5 pts] When can we be certain that the average velocity of an object is always equal to its instantaneous velocity?

        (i) always
        (ii) never
        (iii) only when the velocity is constant
        (iv) only when the acceleration is constant
        (v) only when the acceleration is changing at a constant rate.
      ANS: (iii).
      (b) [5 pts] The figure shows the position of an object (moving along a straight line) as a function of time. Assume two significant figures in each number. Which of the following statements about this object is true over the interval shown?

        (i) The object is accelerating to the left.
        (ii) The object is accelerating to the right.
        (iii) The acceleration of the object is in the same direction as its velocity.
        (iv) The average speed of the object is 1.0 m/s.
      ANS: (i).
      (c) [15 pts] An object is moving in a straight line along the x-axis. A plot of its velocity in the x-direction as a function of time is shown in the figure.

        (i) [5 pts] Which graph below represents its acceleration in the x -direction as a function of time?
        ANS: The left graph.
        (ii) [5 pts] What is the object's displacement in the time interval
        0 � t � 5 s
        ANS: The displacement equals the area under the curve: Dx=11.5 m.
        (iii) [5 pts] What is the object's average acceleration in the time interval
        0 � t � 9 s

        aav
        =
        Dvx
        Dt
        =
        (-2.0 m/s)-(1.0 m/s)
        9.0 s-0
        =
        -0.33 m/s2
  3. Two observers are positioned next to a building of height h, one at the base and the other on the roof. The ground observer launches an object vertically upward with a speed of 16.0 m/s. The second observer stationed on the roof of the building measures the object's speed at 4.00 m/s as it passes by in an upward direction. Neglect air friction.

      (a) [6 pts] How tall is the building?

      v2yf
      =
      v2yi-2g(yf-yi)
      (4.00 m/s)2
      =
      (16.0 m/s)2-2(9.80 m/s2)(yf-yi)
      yf-yi
      =
      12.2 m

      (b) [6 pts] How long did it take the object to reach the top of the building?

      vyf
      =
      vyi-gt
      t
      =
      vyi-vyf
      g
      =
      16.0 m/s-4.00 m/s
      9.80 m/s2
      =
      1.22 s

      (c) [6 pts] If the observer set her clock to zero at the time the object was launched, what did her clock read when the object returned to the ground?

      yf
      =
      yi+vyit-gt2/2
      0
      =
      0+tg[16.0 m/s-(4.90 m/s2)tg]
      tg
      =
      3.27 s

      (d) [7 pts] What velocity did the object have just before hitting the ground?

      vyf = -vyi = -16.0 m/s
  4. A speeder passes by a telephone pole on the side of the highway traveling at 30.0 m/s. A police car parked next to the pole takes off instantly with an acceleration of 5.00 m/s2.

      (a) [9 pts] How long does it take the policeman to catch the speeder?
      The position equations for the speeder and policeman are
      xspeeder
      =
      (30.0 m/s)t
      xpoliceman
      =
      (5.00 m/s2)t2/2
      At t=tm, they meet. Set the two position equations equal to each other.
      (30.0 m/s)tm
      =
      (5.00 m/s2)t2m/2
      tm
      =
      12.0 s

      (b) [8 pts] How far from the telephone pole will the speeder be when caught?

      xm
      =
      (30.0 m/s)(12.0 s)
      =
      360. m

      (c) [7 pts] At the time the policeman caught the speeder, how fast was the policeman's cruiser moving?

      vpoliceman
      =
      at
      =
      (5.00 m/s2)(12.0 s)
      =
      60.0 m/s
FORMULAS
A
=
Ax 
^
i
 
 + Ay 
^
j
 
 + Az 
^
k
 
A
=

�
 
Ax2+Ay2+A2z
 
vx
=
dx
dt
ax
=
dvx
dt
vx,avg
=
vf+vi
2
vx,avg
=
Dx
Dt
ax,avg
=
Dv
Dt
General constant acceleration equations
xf
=
xi+vx,it+axt2/2
vx,f
=
vx,i+axt
v2x,f
=
v2x,i+2axDx
Free fall equations
xf
=
xi+vx,it-gt2/2
vx
=
vx,i-gt
vx,f2
=
vx,i2-2gDx

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On 25 Sep 2015, 11:44.