Name___________________________________
Physics 110 Quiz #1, April 5, 2013
Please show all work, thoughts and/or reasoning in order to receive partial credit. The
quiz is worth 10 points total.
Suppose that an object is moving in a straight line along the positive x-axis. The velocity
of the object as a function of time is shown below.
The velocity of an object as a function of time
25
velocity (m/s)
20
15
10
5
0
0
1
2
3
4
5
6
7
8
9
10
time (s)
1. Over the time interval 0s ≤ t ≤ 5s , the acceleration of the object is
a. positive and increasing with time.
b. positive and constant in time.
c. negative and decreasing in time.
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d. negative and constant in time.
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2. Over the interval of time between 5s ≤ t ≤ 10s , the object is displaced by an amount
a. 0m in the positive x-direction.
b. 50m in the positive x-direction.
c. 100m in the positive x-direction.
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d. 100m in the negative x-direction
3. Suppose that over the interval of time 10s ≤ t ≤ 20s , the object needs to be brought to
rest. What is the least constant acceleration that will bring the object to rest over this
time interval?
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From the information in the problem and from the graph we have,
v fx = v ix + axt → 0 = 20 ms + ax (10s) → ax = −2 sm2 , or 2 sm2 in the negative x-direction.
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4. What is the displacement of the object over the time interval 10s ≤ t ≤ 20s ?
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The displacement is given by (assuming that the initial position of the object at
2
t = 10s is x i = 0m ) x f = x i + v ix t + 12 ax t 2 = (20 ms )(10s) + 12 −2 sm2 (10s) = 100m
(
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)
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5. What is the displacement of the object over the entire time interval 0s ≤ t ≤ 20s ?
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Assuming the initial position is x i1 = 0m , the displacement over the first interval is
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 0 ms + 20 ms 
Δx1
v1 =
→ Δx1 = v1Δt1 = 
(5s) = 50m . The displacement over the second
Δt1
2


Δx
interval is v 2 = 2 →€Δx 2 = v 2Δt 2 = (20 ms )(5s) = 100m . The displacement over the
Δt 2
third time interval is 100m from part 4. The net displacement is the sum of each of
these displacements and is 250m .
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Useful formulas:
Motion in the r = x, y or z-directions
1
2
rf = r0 + v 0r t + ar t
Uniform Circular Motion
Geometry /Algebra
v2
ar =
r
Circles
Triangles
Spheres
v 2 C = 2πr
A = 12 bh
A = 4 πr 2
Fr = mar = m
r
A = πr 2
V = 43 πr 3
2πr
v=
Quadratic equation : ax 2 + bx + c = 0,
T
2
v fr = v 0r + ar t
2
2
v fr = v 0r + 2arΔr
FG = G
m1m2
r2
Vectors
Useful Constants
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2
magnitude of avector = v x + v 2 y
g = 9.8 m s2
v 
direction of avector → φ = tan  y 
 vx 
N A = 6.02×1023 atoms mole
−1
Linear Momentum/Forces
→
→
→
→
→
p f = p i + F Δt
→
→
F = ma
→
G = 6.67×10−11 Nm
2
vsound = 343 m s
Heat
2
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K r = 12 Iω 2
U g = mgh
→
Fs = −k x
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
W R = τθ = ΔE R
W net = W R + WT = ΔE R + ΔE T
ΔE R + ΔE T + ΔU g + ΔU S = 0
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
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Rotational Motion
Fluids
Simple Harmonic Motion/Waves
2π
ω = 2πf =
T
TS = 2π
m
k
TP = 2π
l
g
1
v =±
k  x 2 2
A1− 
m  A2 
x ( t ) = Asin( 2Tπt )
k
cos( 2Tπt )
m
k
a( t ) = −A sin( 2Tπt )
m
FT
v = fλ =
µ
v(t) = A
Sound
v = fλ = (331+ 0.6T) ms
I
; Io = 1×10−12 mW2
I0
v
v
f n = nf1 = n ; f n = nf1 = n
2L
4L
β = 10log
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v
2L
I = 2π 2 f 2 ρvA 2
f n = nf1 = n
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kg 2
kB = 1.38×10−23 J K
σ = 5.67×10−8 W m 2 K 4
Work/Energy
K t = 12 mv
p = mv
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−b ± b 2 − 4ac
2a
whose solutions are given by : x =