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Ch 2 Standardized Test Prep, pg 74
1] B [2] F [3] D [4] H [5] B [6] G [7] C [8] G
[9] D
10] Displacement measured only the net change in position from starting point to
end point. The distance traveled is the total length of the path followed from
starting point to end point and may be greater than or equal to the
displacement.
11a] ΔX1 = +2400 m; V1 = +4.0 m/s
b] ΔX2 = +1500 m; V2 = +2.5 m/s
c] ΔX3 = +900 m; V3 = +2 m/s
d] ΔXtot = +4800 m; Vavg = 2.7 m/s
1 A bus travels 270 km south along a straight path with
an average velocity of 90 km/h to the south. The bus stops for 30 min. Then,
it travels 320 km south with an average velocity of 80 km/h south.
a] How long does the total trip last? A6, pg44
Given/Find
ΔX1 = 270 km, S ΔX1 = V1●Δt1
ΔX2 = V2●Δt2
V1 = 90 Km/h 270 = 90Δt1
320 = 80Δt2
Stop = 30 min = 0.5 h 270/90 320/80
ΔX2 = 320 km, S 3 h = Δt1
4 h = Δt2
V2 = 80 km/h
Total Δt = ? Total
Δt
= Δt1 +
Δt2 + Stop
Total Δt
= 3 + 4 + 0.5 = 7.5 h
b] What is the average velocity for the total trip?
Vavg = Total Displacement = 270 + 320 km = 590 km = 78.6
km/h, S
Total Time 7.5 h 7.5 h
2 A treadmill runs with a velocity of -2.0 m/s and
speeds up at regular intervals during a half-hour workout. After 20 min, the
treadmill has a velocity of -6 m/s.
What is its average acceleration during this period? B4, pg 49
Given/Find
Vi = -2.0 m/s a = Vf - Vi = -6.0 -(-2.0) =
Vf = -6.0 m/s Δt 1200 s
Δt = 20 min x 60 s/ min = 1200 s
a = ? a = - 4.0 = -3.33 x10-3
m/s2
1.2x103
3 Suppose a treadmill has an average acceleration of 6.0
x10-3 m/s2.
a] How much does its speed change after 5.0 min? B5, pg 49
Given/Find
a = 6.0 x10-3 m/s2.
a = ΔV ----> ΔV = a Δt
Δt = 5 min x 60 s/ min = 300 s Δt = 6.0 x10-3
x 300
ΔV = ? ΔV = 1.8 m/s
b] If the treadmill’s speed is 2.0 m/s, what will its final speed be?
Vf = ? ΔV = Vf - Vi
1.8 = Vf - 2
+2.0 + 2
3.8 = Vf
4 A car enters the freeway with a speed of 6.0 m/s and
accelerates uniformly for 3.0 km in 4.0 min.
How fast (m/s) is the car moving after this time? C4, pg 53
Given/Find
Vi = 6.0 m/s
ΔX = 3.0 km x 1000 m/km = 3000 m
Δt = 4.0 m x 60 s/min = 240 s
Vf = ?
ΔX = (Vf + Vi)Δt --> 3000 = (Vf + 6)240 ---> 3000 = (Vf +6)120
2 2
3000/120 = Vf + 6
25 = Vf + 6
19 = Vf
5 A driver of a car traveling at 16.0 m/s applies the
brakes, causing a uniform acceleration of -2 m/s2.
a] How long does it take the car to reach 10 m/s? D4, pg 55
Given/Find
Vi = 16.0 m/s a = Vf - Vi --> Δt = Vf - Vi
Vf = 10 m/s Δt a
a = -2 m/s2
Δt = ? Δt = 10 - 16 = -6 = 3.0 s
-2 -2
b] How far has the car moved during the braking period?
ΔX = ? ΔX = (Vf + Vi)Δt
2
ΔX = (10 + 16)3 = (26)3 = 39
m
2 2
6 A car traveling initially at +5.0 m/s accelerates
uniformly at the rate of 3.0 m/s2 for a distance of 200 m.
What is its velocity at the end of the acceleration? E2, pg 58
Given/Find
Vi = 5.0 m/s a = Vf - Vi ΔX = (Vf + Vi)Δt
Vf = ? m/s Δt 2
a = 3.0 m/s2
ΔX = 200 m 3.0 = Vf - 5 200 = (Vf + 5) Δt
Δt 2
Δt = Vf - 5 ---> 200 = (Vf + 5)(Vf - 5)
3.0 2 3.0
200 = (Vf2 - 25)
6.0
6(200) = Vf2 - 25
1200 = Vf2 - 25
+25 + 25
1225 = Vf2
√1225 = Vf
35 m/s = Vf
7 A car accelerates uniformly in a straight line from
rest at 3 m/s2.
a] What is the speed of the car after it has traveled 60 m? E3, pg 58
Given/Find
Vi = 0.0 m/s a = Vf - Vi --> Δt = Vf - Vi ΔX = (Vf +
Vi)Δt
Vf = ? m/s Δt a 2
a = 3 m/s2
ΔX = 60 m Δt = Vf - 0 = Vf 60 = (Vf)
Vf
3 3 2 3
60 = Vf2
6
360 = Vf2
√360 = Vf
19.0 m/s = Vf
b] How long did it take?
Δt = ? Δt = Vf/3 = 19.0/3.0 = 6.33 s
8 An aircraft has a liftoff speed of 60 m/s. What
minimum constant acceleration does this require if the aircraft is to be
airborne after a take-off run of
450 m?
E5, pg 58
Given/Find
Vi = 0.0 m/s ΔX = (Vf + Vi)Δt a = Vf - Vi
Vf = 60 m/s 2 Δt
a = ? m/s2
ΔX = 450 m 450 = (60 + 0) Δt a = 60 - 0
2 15
450 = 30Δt
450/30 = Δt a = 4 m/s2
15 s = Δt
9 A pot falls from a windowsill 20.0 m above the
sidewalk.
a] Draw & label the diagram.
b] How fast is it moving when it hits the ground? F2
Given/Find
Vi = 0.0 m/s a = Vf - Vi ΔX = (Vf + Vi)Δt
Vf = ? m/s Δt 2
a = 9.8 m/s2
ΔX = 20 m 9.8 = Vf - 0 20 = (Vf + 0) Δt
Δt 2
Δt = Vf 20 = (Vf) Vf
9.8 2 9.8
Vf2
20 = -----
19.6
19.6*20 = Vf2
_______
19.8 m/s = Vf <--- √19.6*20 = Vf
c] How much time doe it take?
Δt = ? Δt = Vf = 19.8 = 2.02 s
9.8 9.8
10 Starting at 2.0 m above the floor a volley ball is hit
straight up with an initial velocity of +5.0 m/s.
a] Draw & label the diagram.
F4,pg 64
/|\ +
V top = 0 m/s
/|\ Vf | = 1.0 m/s
Y2 |-------+
Imagine a parabola connecting the
+
|
Vi | = 5.0 m/s
|
Y1 +----------------------------------------------+
| |
| | -Vi
| |
| \|/
Y0 -----------------------------------------------
//////////////////////Floor\\\\\\\\\\\\\\\\\\\\\\
b] Find the displacement of a volley ball when its velocity is
+1 m/s upward.
Given/Find
Vi = 5.0 m/s a = Vf - Vi
Vf = 1.0 m/s at top Δt
a = -9.8 m/s2
ΔY = ? m -9.8 = 1.0 - 5.0 --->Δt = -4.0 = 0.408 s
Δtup
-9.8 &nbssp;
ΔY = (Vf + Vi)Δt --->ΔY = (1.0 + 5.0) (0.408 s)
2 2
ΔY = 1.22 m above launch point,
c] Find its position. F4
Given/Find
ΔY = 1.22 m ΔY = Y2 -Y1
Y1 = 2m 1.22 m = Y2 - 2 .0 m
y2 = ? +2.00 m + 2.0 m
3.22 m = Y2 ------>Y2 = 3.22 m above floor
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