Here You Go

Only available on StudyMode
  • Download(s) : 95
  • Published : December 15, 2012
Open Document
Text Preview
Chapter 4

11. Walker3 4.P.050. [544504] Show DetailsA train moving with a constant velocity travels 120 m north in 12 s and an undetermined distance to the west. The speed of the train is 26 m/s. (a) Find the direction of the train's motion relative to north.[pic]67.4° west of north. (b) How far west has the train traveled in this time? [pic]288 m

[Answer]

Distance traveled = 26m/s *12s = 312 m
Displacement along north = 120m

a) Angle relative to north = arccos ( 120m/312m) = 67.4°
b) Displacement along west = 312m * sin (67.4°) = 288 m

12. Walker3 4.P.051. [544514] Show Details A hummingbird is flying in such a way that it is initially moving vertically with a speed of 5.1 m/s and accelerating horizontally at 11.0 m/s2. Assume the bird's acceleration remains constant for the time interval of interest. [pic]

(a) Find the x component of the hummingbird's velocity at the time t = 0.45 s. [pic]4.95 m/s (b) Find the y component of the hummingbird's velocity at the time t = 0.45 s. [pic]5.1 m/s (c) What is the bird's direction of travel at this time relative to the positive x-axis?. [pic]45.9°

[Answer]

(a) Motion in x-direction starts from rest and has constant acceleration. Therefore, x-component of hummingbird’s velocity is = v0+a*t = 0+11.0*0.45=4.95m/s (b) y-component = v0+a*t = 5.1 + 0*0.45 =5.1m/s (comments: well, it is also reasonable to assume a=g=-9.81m/s2). (c) Angle relative to x-axis = arctan ( vy/vx) = 45.9°

13. Walker3 4.P.053. [544631] Show Details A hot air balloon rises from the ground with a velocity of (2.2 m/s)[pic]. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (6.0 m/s)[pic] relative to the balloon. When opened, the bottle is 5.8 m above the ground. (Neglect air resistance.) (a) What is the initial velocity of the cork, as seen by an observer on the ground? [pic] = ( [pic]6 m/s)[pic] + ( [pic]2.2 m/s)[pic] (b) What is the speed of the cork, and its initial direction of motion, as seen by the same observer? |Speed |[pic]6.39 m/s |

|Direction |[pic]20.1° above horizontal |

c) Determine the maximum height above the ground attained by the cork. [pic]6.05 m (d) How long does the cork remain in the air? [pic]1.33 s

[Answer]
a) Vg = Vballon + Vcork = 2.2m/s [pic] + 6.0m/s [pic]
b) Speed = sqrt (2.2m/s ^2 + 6.0m/s ^2) = 6.39m/s
Angle relative to x-axis = arctan ( 2.2 / 6.0 ) = 20.1° above horizontal c) yf = yi + vy,0*t – 1/2 g * t2
at maximum height, vy = 0, so the
yf = yi + vy,02/ (2*g) = 5.8 + 2.22/(2*9.81) = 6.05m
d) When the cork reaches the ground, yf = 0.0 = 5.8 + 2.2*t – 4.905*t2 , therefore t = ( 2.2 + sqrt(2.22-4*4.905*(-5.8))) / (2*4.905) = 1.33 s

14. Walker3 4.P.057. [544672] Show Details When the dried up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.6 m/s at an angle of 60.0° above the horizontal. [pic]

(a) If the seed pod is 1.2 m above the ground, how long does it take for the seed to land? (Neglect air resistance.) [pic]0.775 s (b) What horizontal distance does it cover during its flight? [pic]1.01 m

[Answer]
a) Similar to problem 13.
yf = 0.0 = yi + vy,0*t – 1/2*g *t2 = 1.2 + (2.6*sin(60.0°)) * t – 4.905 *t2 , t must be positive, therefore t = 0.775 by solving the above the 2nd order equation

b) xf - xi = vx,0*t = (2.6*sin(30.0°))*0.775 = 1.01 m

15. Walker3 4.P.065. [544703] Show Details The men's world record for the shot put, 23.12 m, was set by Randy Barnes of the United States on May 20, 1990. If the shot was launched from 5.70 ft above the ground at an initial angle of 44.0°, what was its initial speed? [pic]14.5 m/s

[Answer]
Motion in y-axis
0 = 5.70ft + v * sin44° * t – 4.905*t2, therefore t = 1.59 s Motion in x-axis
v*cos44° * t = 23.12 m
Combined the above 2 equations
=> v = 14.5m/s

16. Walker3 4.P.072. [544644] Show...
tracking img