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physics 105 fall 2006 solutions 1A
g 9.8
1. In May 1998, forest fires in southern Mexico and Guatemala spread smoke all the way to Austin. Those fires
consumed forest land at a rate of 23100 v 23100 acres/week. How many square meters of forest are burned
down every minute?.
(1 acre = 4.84 103 yd2 acre 60.5 80 , 1 yd = 36 yd 36 inch, 1 inch = 2.54 cm. )
A) 4072
B) 5463
C) 6826
D) 8047
E) 9274
1E
2
2
v  acre  yd  in
3
1. acre w l week 7 24 60 minutes v
v  9.274 10
week
2.
The information on a one-gallon paint can is that the coverage, when properly applied, is 450 ft2 One
gallon is 231 in3 . What is the average thickness of the paint in such an application?
A) 0.0036 in B) 0.043 in
C) 0.090 in
D) 0.21 in
E) 0.51 in
2A
gal
3
d
a ft 12
d  3.565 10
2

A ft
2
3. A particle moves according to the equation x = 5 - 12.t + 3.t2 x( t ) 5 12 t 3 t
where x is in meters
and t is in seconds. Find the position of the particle at t = 2 seconds.
A) 0
B) 3
C) 6
D) -12
E) -7
x( 2)  7
4. An automobile travels on a straight road for 40 km at 20 km/hr. It then continues in the same direction for
another 40 km at 80 km/hr v2 80. What is the average velocity of the car during this 80 km trip?
A) 32 B) 36 C) 42 D) 46 E) 50
L
L
time to make first 40 km distance t1
time to make the second 40 km distance t2
total distance = 2L
v1
v2
2L
total time = t1 + t2 vavg
vavg  32
t1 t2
5.
Add the two vectors shown. The magnitude of the result is:
A) 10.9 B) 16.9 C) 28.7 D) 32.9 E) 36 5B
 37 deg
Sum
21 j ( 15 cos (  )  i 15 sin(  )  j)
Sum  16.937
  


A = 2i points east and vector B = -5j points south. If vector C = B – A ,

then vector C points:
6. Vector
A) - 680 B) 680
A
2 i
B
C) -1120 D) +1120 E) none of these
2
5 j
C B A C
angle( C i)  111.801
5
7. A train slowly climbs a 500-m mountain track which is at an angle of 100 with respect to the horizontal. How
much altitude does it gain?
h L sin(  )
A) 86.8 m B) 88.2 m C) 341 m D) 492 m E) 50.7 m 7A
h  86.824
8. Find the resultant of the following two vectors: i) 50 units due east A 50 i and ii) 100 units 300 north of
west B 150 deg . vector B = B 100 cos ( B)  i 100 sin( B)  j
A) 100 units 300 north of west B) 62 units 150 north of west C) 87 units 600 north of west D) 62 units 540
north of west E) 126 units 150 north of east
8D
C
A
B
C
36.603
50
C  61.966
angle( C i)  126.206
9. A velocity vs time plot of a runner is shown in the figure.
What is the
distance he made between 2 and 6 seconds?
distance = 2*0.6+2*0.3
10. A ball is rolled horizontally off a table with an initial speed
of 0.24 m/s. A stopwatch measures the ball’s trajectory time
from table to the floor to be 0.30 s. What is the height of the table? (g = 9.8 m/s 2 and air resistance is negligible)
1  2
A) 0.11 m B) 0.22 m C) 0.33 m D) 0.44 m E) 0.55 m
10D
g t  0.441
2
11. A track star in the broad jump goes into the jump at 12 m/s vo 12 and launches himself at 200 above the horizontal

20 deg . How long is he in the air before returning to Earth? (g = 9.8 m/s 2)
2 vo  sin(  )
A) 0.42 s B) 0.84 s C) 1.25 s D) 1.48 s E) 1.68
11B
 0.838
g
12. A ball is launched from ground level at 30 m/s vo 30 at an angle of 350  35 deg above the horizontal. How far
does it go before it is at ground level again?
A) 14 m B) 21 m C) 43 m D) 86 m E) 106 m 12D
R
2
vo  sin ( 2   )
R  86.298
g
Workout problems:
Workout problem 1 - 3 points
Oasis B is 25 km due east of oasis A. Starting from oasis A, a camel walks 24 km in a direction 15 0 south of
east 
15 deg . How far is the camel then from oasis B?
Oasis B location is at B 25 i 0 j first displacement vector from oasis A is A 24 cos (  )  i 24 sin(  )  j
The camels should walk first a vector A (known) then a vector x i y  j (unknown) to finish at oasis B (location
known)
we can write that A ( x i y  j) B so
[24cos()+x]i + 24sin()+y]j = 25i + 0j
equations for x and y components are respectively 24 cos (  ) x 25 and 24 sin(  ) y 0
x 25 24 cos (  )
y
24 sin(  )
Workout problem 2 - 2 points
x  1.818
2. A jumbo jet must reach a speed of 400 km/hr v
y  6.212
400 
1000
and the distance is
2
x
2
y  6.472
m/s on the runway for takeoff.
3600
a) What is the least constant acceleration needed for takeoff from a 2 km x 2000 runway?
b) How much time it would take to cover the distance of 2 km?
a)
v
b)
x
2
2
0
1  2
at
2
2
2 a  x >>
>>
a
t
v
2x
2x
a
a  3.086
t  36