(kma786) armington hw0201 ete (57165) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points If you drop an object, it will accelerate downward at a rate of g = 9.8 m/s2 . If you throw it downward instead, its acceleration (in the absence of air resistance) will be 1. less than 9.8 m/s2 . 2. 9.8 m/s2 correct 3. greater than 9.8 m/s2 . 4. Unable to determine. Explanation: The acceleration due to the gravity is independent of any initial velocity and remains constant. 002 10.0 points A stone is thrown straight upward and at the top of its trajectory its velocity is momentarily zero.
1
The gravitational acceleration near the surface of the earth is considered constant for all practical purposes. This acceleration of 9.8 m/s2 points downward. 003 10.0 points A basketball player achieves a hang time of 0.94 s in dunking the ball. What vertical height will he attain? The acceleration of gravity is 9.8 m/s2 . Correct answer: 1.08241 m. Explanation: Let : t = 0.94 s and g = 9.8 m/s2 .
t It takes the basketball player t = to 2 rise to his maximum height, dened by g t2 g t2 1 = h = g (t)2 = 2 24 8 (9.8 m/s2 ) (0.94 s)2 = 1.08241 m . = 8 004 10.0 points The acceleration of a marble in a certain uid is proportional to the speed of the marble squared, given by a = v2 , where v > 0 m/s and = 1.61 m1 . If the marble enters this uid with a speed of 0.96 m/s, how long will it take before the marbles speed is reduced to half of its initial value? Correct answer: 0.646998 s. Explanation:
What is its acceleration at this point? 1. 9.8 m/s down correct 2. Unable to determine 3. 9.8 m/s2 up 4. Zero Explanation:
2
Let :
= 1.61 m1 v0 = 0.96 m/s . a= dv = v2 dt dv = dt v2 dv = dt . v2
and
armington (kma786) hw0201 ete (57165) The marbles speed is reduced to half of its initial value, so
v0 /2 v0
2
Explanation:
t
vt =
0
a dt is the area between the accel-
u2 du =
0
t
dt
t
1 v
v0 /2 v0
= t
0
1 2 v0 v0
= (t 0)
1 = t v0
eration curve and the t axis during the time period from 0 to t. If the area is above the horizontal axis, it is positive; otherwise, it is negative. In order for the velocity to be zero at any given time t, there would have to be equal amounts of positive and negative area between 0 and t. According to the graph, this condition is never satised. 006 (part 1 of 4) 10.0 points Consider the plot describing motion along a straight line with an initial position of 10 m. 6 5 velocity (m/s) 4 3 2 1 0 1 2 3 4567 time (s) What is the velocity at 2 seconds? Correct answer: 5 m/s. 4 1 2 3 8 9
1 1 = t= 1 ) (0.96 m/s) v0 (1.61 m = 0.646998 s .
005 10.0 points Consider the acceleration of an object starting from rest. 5 4 Acceleration (m/s2 ) 3 2 1 0 1 2 3 4 5
2 3 4 5 Time (s) Other than at t = 0, when is the velocity of the object zero? 1. At no other time on this graph correct 2. 4.0 s 3. 3.5 s 4. 5.0 s 5. During the interval from 1.0 s to 3.0 s
0
1
Explanation: The velocity at 2 seconds can be read from the plot; however, it can also be calculated: a= v 5 m/ s 0 m/ s = t 2s0s 2 = 2.5 m/s and
v = vi + a (tf ti ) = 5 m/ s .
= 0 m/s + (2.5 m/s2 ) (2 s 0 s)
armington (kma786) hw0201 ete (57165) 007 (part 2 of 4) 10.0 points What is the position at 2 seconds? Correct answer: 15 m. Explanation: Let : 6 5 4 velocity (m/s) 3 2 1 0 1 2 3 456 time (s) 7 8 9 x = xi + vi (tf ti ) + 1 a ( tf ti ) 2 2 = 15 m + (5 m/s) (6 s 2 s) 1 + (0.25 m/s2 )(6 s 2 s)2 2 = 37 m . x0 = 10 m . = 37 m ; however can it also be calculated a= v 6 m/ s 5 m/ s = t 6s2s 2 = 0.25 m/s and
3
The position is 15 m plus the area of the trapezoid from 2 seconds to 6 seconds x = 15 m + 1 (6 s 2 s)(6 m/s + 5 m/s) 2
1 2 3 4
009 (part 4 of 4) 10.0 points What is the position at 9 seconds? Correct answer: 31 m. Explanation: The position is 37 m plus the area of the triangle x = 37 m + = 31 m ; however, it can also be calculated: a= v 4 m/s 0 m/s = t 9s6s = 1.33333 m/s2 and 1 (9 s 6 s)(4 m/s 0 m/s) 2
The position at 2 seconds is 10 m plus the area of the triangle (shaded above) x = 10 m + = 15 m ; however, it can also be calculated: 1 x = xi + vi (tf ti ) + a (tf ti )2 2 = 10 m + (0 m/s) (2 s 0 s) 1 + (2.5 m/s2 )(2 s 0 s)2 2 = 15 m . 1 (2 s 0 s)(5 m/s 0 m/s) 2
x = xi + vi (tf ti ) +
008 (part 3 of 4) 10.0 points What is the position at 6 seconds? Correct answer: 37 m. Explanation:
1 a ( tf ti ) 2 2 = 37 m + (0 m/s) (9 s 6 s) 1 + (1.33333 m/s2 )(9 s 6 s)2 2 = 31 m . 010 10.0 points
armington (kma786) hw0201 ete (57165) An ant starts at one edge of a long strip of paper that is 40 cm wide. She travels at 2.5 cm/s at an angle of 30 with the long edge. How long will it take her to get across? Correct answer: 32 s. Explanation: Let : v = 2.5 cm/s , d = 40 cm , and = 30 . = 180 96 = 84 , so applying the law of cosines, a r
4
b
d
v
Although her velocity is 2.5 cm/s, she advances toward her goal (the opposite edge) at less than this speed. In fact, her actual speed is the hypotenuse of a right triangle, and the rate at which she approaches the opposite edge is v = v sin , so t= d 40 cm = v (2.5 cm/s) sin 30
r 2 = a2 + b2 2 a b cos = (15.8)2 + (6.7)2 2 (15.8) (6.7) cos 84 = 272.399 r = 272.399 = 16.5045 .
012 (part 2 of 2) 10.0 points Find the angle between the direction of the resultant vector A + B and the direction of the vector A. Correct answer: 23.8113 . Explanation:
= 32 s . 011 (part 1 of 2) 10.0 points Consider two vectors A and B and their resultant A + B . The magnitudes of the vectors A and B are, respectively, 15.8 and 6.7 and they act at 96 to each other.
r a
b
Applying the law of sines, B A+B r b = sin sin b sin sin = r = arcsin = arcsin a = 15.8 , b = 6.7 , and = 96 . b sin r 6.7 sin 84 16.5045
A Find the magnitude of the resultant vector A + B. Correct answer: 16.5045. Explanation: Let :
= 23.8113 .
013 (part 1 of 2) 10.0 points
armington (kma786) hw0201 ete (57165) Express the vector R
5
A
P
C R D
6. P = B A 7. P = C + B 8. P = A B correct 9. P = A D 10. P = B + A Explanation: By the triangle method of addition B+P =A P =AB. 015 10.0 points A car travels 2.82 km in the x-direction, then turns left 48.2 to the original direction and travels an additional distance of 1.37 km. Calculate the x component of the cars net displacement. Let : d1 = 2.82 km , = 48.2 , and d2 = 1.37 km .
B in terms of A, B , C , and D , the edges of a parallelogram.
1. R = A C 2. R = A B 3. R = A + D 4. R = B + A correct 5. R = B A 6. R = A D 7. R = C + B 8. R = C + D 9. R = D A 10. R = B + D Explanation: Apply the parallelogram rule of addition: join the tails of the two vectors A and B ; the resultant vector is the diagonal of a parallelogram formed with A and B as two of its sides. 014 (part 2 of 2) 10.0 points Express the vector P in terms of A, B , C , and D, 1. P = A + D 2. P = C + D 3. P = B + D 4. P = D A 5. P = C A
Correct answer: 3.73315 km. Explanation: The x component of the cars net displacement is simply dx = d1 + d2 cos = 2.82 km + (1.37 km) cos 48.2 = 3.73315 km .