The eye of a hurricane passes over New Orleans in a direction 60. 0° north of west with a speed of 41. 0 km/hr. Three hours later, the hurricane suddenly shifts due north, and its speed slows to 25. 0 km/hr. How far from New Orleans is the hurricane 4. 50 hours after it passes over the city? v1 = 41. 0km/hr(cos30°, sin30°) = (35. 5, 20. 5) km/hr v2 = (0, 25. 0)km/hr ? D = v1×3hrs + v2 × 1. 5hrs =Answer (105. 9, 99) km
A ship which can move at the velocity of v0 = (12, 6) m/s (x-axis: the West to the East) on the lake.
The ship is now on the river which flows to the West at the speed of 5 m/s. 12 points) a. Calculate the velocity of the ship on the river. vwater = (-5, 0)m/s, v = v0 + vwater =Answer (7, 6) m/s b. Calculate the speed of the ship on the river. |v| =Answer 9. 22 m/s c. The ship started sailing at A which is 23 km west and 47 km south from the origin (0. 0). Find the location of the ship after sailing 5 hours.
x = A + vt = (-23km, -47km) + (7, 6)m/s × 5 × 3600s = (-23km, -47km) + (126, 108)km = Answer (103, 61) km d. How far did the ship sail in the distance from the initial location. |D| = Answer 166. 0km
A Boeing 747 airplane is flying at the elevation of 2,000 km with the constant speed of 250 m/s around 300 km away from Baton Rouge. a. How long does the airplane take to reach Baton Rouge with this speed? t = 300km / 250 m/s = 1200s = Answer 20 minutes. b. Near Baton Rouge, the airplane needs to slow down to the landing speed of 150 m/s before it starts the landing gear. How long does the airplane take to get the landing speed, if it slows down with the constant acceleration of -2 m/s2 in the horizontal direction? 150m/s = 250m/s – 2m/s2 t ? t = Answer 50s. . While the airplane is slowing down in the problem b) above, how long horizontal distance does this airplane fly? This calculation will give the distance from the airport where the airplane starts preparing landing. Distance = 250m/s×50s + ½ (-2m/s2) (50s)2 = 10000m = Answer 10. 0 km. d. While the airplane is slowing down in the problem b) above, if the airplane lowers its elevation by free-falling, what would be the elevation before the landing gear? y = y0 + ½gt2 = 2000km + ½ (-9. 8m/s2) (50s)2 = 2000 km – 12. 25 km = Answer 1987. 75 km. e.
By using the results from b) and c), what is the actual distance including both horizontal and vertical distances that the airplane flies during this duration. In other words, what is the magnitude of the displacement vector for this duration. displacement = v((10. 0 km)2 + (-12. 25 km)2) = Answer 15. 81 km. f. What is the angle of flight with respect to the horizontal plane for this duration. In other words, what is the direction of the displacement vector for this duration. ? = tan-1[(-12. 25km)/(10. 0 km)] = -50. 77° = Answer 0. 886 radians g. When the airplane lands on the ground, its velocity is 50 m/s.
The length of the runway of the Baton Rouge airport (BTR) is 2 km or 2000 m. If the airplane slows down with a constant acceleration, what would be the minimum acceleration needed for this airplane to stop within the runway? Acceleration = [0 – (50m/s)2)] / (2×2000m) = Answer – 0. 625m/s^2. A displacement vector r has a magnitude of r = 175 m and points at an angle 50 relative to the x axis. Find the x and y components of this vector Y= r sin = (175m)(sin 50) = Answer 134 m X = r cos (175m) (cos 50) = Answer 112 m A jogger runs 145 m in a direction 20. ° east of north (displacement vector A) and then 105 m in a direction 35. 0° south of east (displacement vector B).
Using components, determine the magnitude and direction of the resultant vector C for these two displacements Ax= (145 m) sin 20. 0°= 49. 6 m Ay= (145 m) cos 20. 0°= 136 m Bx = (105 m) cos 35. 0° = 86. 0 m By= -(105 m) sin 35. 0° = 60. 2 m Note that the component By is negative, because y points downward, in the negative y direction in the drawing. Substituting these values into the results for C and @ gives C= ?(Ax+Bx)^2+(Ay+By)^2= ?(49. 6+86)^2+(136+60. )^2= Answer 155 m @= tan^-1(Ay+By/Ax+Bx)= tan^-1 (136-60. 2)/(49. 6+86. 0)= Answer 29 degrees Two displacement vectors A and B . Vector A points at an angle of 22. 0° above the x axis and has an unknown magnitude. Vector B has an x component Bx = 35. 0 m and has an unknown y component By . These two vectors are equal. Find the magnitude of A and the value of By . Concept Questions and Answers What does the fact that vector A equals vector B imply about the magnitudes and directions of the vectors? Answer When two vectors are equal, each has the same magnitude and each has the same direction.
What does the fact that vector A equals vector B imply about the x and y components of the vectors? Answer When two vectors are equal, the x component of vector A equals the x component of vector B (Ax = Bx) and the y component of vector A equals the y component of vector B (Ay = By). The x components of the vectors are the same and the y components of the vectors are the same. A cos 22. 0 degrees (Component Ax of vector A) = 35. 0 m (component Bx of vector B) A sin 22. 0 degrees (Comp Ay of vector A) = By (comp By of vector B) A sin 22/A cos 22. 0 = By/35. 0 m Value of By = (35. 0)sin 22/cos 22 = (35. m) tan 22. 0 = Answer 14. 1 m Magnitude of A = 35. 0 m/cos 22. 0 = Answer 37. 7 m Can two nonzero perpendicular vectors be added together so their sum is zero? No Can three or more vectors with unequal magnitudes be added together so their sum is zero? Yes Vectors A, B, and C satisfy the vector equation A+B=0 , and their magnitudes are related by the scalar equation A^2+B^2=C^2. How is vector A oriented with respect to vector B? Answer: Vector A is perpendicular to Vector B Vectors A, B, and C satisfy the vector equation A+B=C, and their magnitudes are related by the scalar equation A +B=C.
How is vector A oriented with respect to vector B? Answer: Vector A points in same direction as Vector B Which of the following displacement vectors (if any) are equal? Variable A Magnitude 100 m Direction 30 degree N of E Variable B Magnitude 100 m Direction 30 deg S of W Variable C Magnitude 50 m Direction 30 deg S of W Variable D Magnitude 100 m Direction 60 deg E of N Answers: Variable A and D Are two vectors with the same magnitude necessarily equal? Answer= No The tail of a vector is fixed to the origin of an x, y axis system. Originally the vector points along the + x axis and has a magnitude of 12 units.
As time passes, the vector rotates counterclockwise. What are the sizes of the x and y components of the vector for the following rotational angles? (a) 90deg (b)180deg (c) 270deg (d) 360deg Answer: Ax= 0 units Ay= +12 units B) Ax= -12 units Ay= 0 units, C) Ax=0 units Ay= -12 units, D) Ax=+12 units Ay= 0 units Which are vectors or scalars? Position is a vector Distance is a scalar Displacement is a vector Speed is a scalar Velocity is a vector Acceleration is a Vector Mass is a scalar Force is a vector Weight is a vector Gravitational acceleration is a vector The drawing shows projectile motion at hree points along the trajectory. The speeds at the points are v1, v2, and v3. Assume there is no air resistance and rank the speeds, largest to smallest. Answer is (a) v1> v>3> v2 (b) v1 > v2 > v3 (c) v2 >v3> v1 (d) v2 > v1> v3 (e) v3 >v2> v1 Two balls are thrown from the top of a building, as in the drawing. Ball 1 is thrown straight down, and ball 2 is thrown with the same speed, but upward at an angle _ with respect to the horizontal. Consider the motion of the balls after they are released. Which one of the following statements is true Answer: (c) Both balls have the same acceleration at all times.
Two objects are fired into the air, and the drawing shows the projectile motions. Projectile 1 reaches the greater height, but projectile 2 has the greater range. Which one is in the air for the greater amount of time? Answer (a) Projectile 1, because it travels higher than projectile 2. Your car is traveling behind a jeep. Both are moving at the same speed, so the velocity of the jeep relative to you is zero. A spare tire is strapped to the back of the jeep. Suddenly the strap breaks, and the tire falls off the jeep. Will your car hit the spare tire before the tire hits the road? Assume that air resistance is absent
Answer (b) No. The car will not hit the tire before the tire hits the ground, no matter how close you are to the jeep because velocity of falling tire is equal to velocity of car In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0. 050 s, during which time it experiences an acceleration of 340 m/s2. The ball is launched at an angle of 51° above the ground. Determine the horizontal and vertical components of the launch velocity. Ax= a [email protected]=340 m/s^2(cos51)=213. 97 m/s^2 Ay= a [email protected]=340 m/s^2(sin51)=264. 23 m/s^2 Vx= (ax)(t)=213. 97(0. 050s)= Answer 10. 0 m/s Vy= (ay)(t)=264. 23(0. 050s)= Answer 13. 21 m/s If the football was placed 30 m (32. 8 yards) away from the goal a) how long does the ball take to get to the goal? Y=Voy(t) + ½ ay(t)^2 T= v2 x 30m/9. 8=Answer 2. 47 s
A mountain-climbing expedition establishes two intermediate camps, labeled A and B in the drawing, above the base camp. What is the magnitude _triangle r of the displacement between camp A and camp B? side AC= (19,600-11,200)m= 8400 m Side BC= (4900-3200m)=1700 m AB=vAC^2+BC^2=v(8400)^2+(1700)^2= Answer 8570. 3 m Suppose that the plane in Example 3 is traveling with twice he horizontal velocity—that is, with a velocity of _230 m/s. If all other factors remain the same, determine the time required for the package to hit the ground. Y= Voy(t)+ ½ ay(t)^2 T=v2 x (-1050)/(-9. 8)= Answer 14. 6 s Michael Jordan, formerly of the Chicago Bulls basketball team, had some fanatic fans. They claimed that he was able to jump and remain in the air for two full seconds from launch to landing. Evaluate this claim by calculating the maximum height that such a jump would attain. For comparison, Jordan’s maximum jump height has been estimated at about one meter. T=20 s Tdown = t/2 Y=yo + Voy(t)+ ½ ay(t)^2
Yo=h Voy=0 m/s Ay= -9. 8 m/s^2 Y= h+ ½ ay(t)^2 H=(-1/8) x (-9. 8)^2(2. 0)^2= Answer 4. 9 m Two cars, A and B, are traveling in the same direction, although car A is 186 m behind car B. The speed of A is 24. 4 m/s, and the speed of B is 18. 6 m/s. How much time does it take for A to catch B? t=d/v = (186m)/(5. 8 m/s)= Answer 32 s A swimmer, capable of swimming at a speed of 1. 4 m/s in still water (i. e. , the swimmer can swim with a speed of 1. 4 m/s relative to the water), starts to swim directly across a 2. 8-km-wide river. However, the current is 0. 91 m/s, and it carries the swimmer downstream. a) How long does it take the swimmer to cross the river? T=width/Vswimmer= (2. 8 km)/(1. 4 m/s)= Answer 2 x 10^3 s (b) How far downstream will the swimmer be upon reaching the other side of the river? X=(Vwr) x (t)= (0. 91 m/s) x (2. 0 x 10^3s) = Answer 1. 8 x 10^3 m A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 5. 0 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 25° when they move past the window, as the drawing shows. How fast is the train moving? Vrt=Vrg/cos 25 = 5/cos 25 Vrg= -Vrt = -5/cos25
Vrg=v(-5/cos25)^2- (5)^2= Answer 2. 3 m/s The x components of the craft’s initial velocity and acceleration are v0x= +22 m/s and Ax= 24 m/s2, respectively. The corresponding y components are v0y=14 m/s and ay=12 m/s2. At a time of t =7. 0 s, find the spacecraft’s final velocity (magnitude and direction). V= v(Vox + axt)^2 + (Voy + ayt)^2= v[(22m/s)+(24m/s^2)(7. 0s)]^2+ [(14m/s)+(12m/s^2)(7. 0s)]^2= ANSWER= 210 m/s @= tan^-1(Voy +ayt)/(Vox + axt)= tan^-1[114m/s + 12 m/s ^2(7. 0s)]/[22 m/s + 24 m/s^2(7. 0s)= ANSWER 27 degrees an airplane moving horizontally with a constant velocity of _115 m/s at an altitude of 1050 m.
The directions to the right and upward have been chosen as the positive directions. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground. T=v2y/ay=v2(-1050)/(-9. 8 m/s^2)= ANSWER= 14. 6 s a care package falling from a plane, and Figure 3. 9 shows this package as package B. As in Example 3, the directions to the right and upward are chosen as the positive directions, and the plane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050 m.
Ignoring air esistance, find the magnitude v and the directional angle @ of the final velocity vector that the package has just before it strikes the ground. V=vvox)^2+(voy)^2+2ay(y)= v(115m/s)^2+(0m/s)^2+2(-9. 8m/s^2)((-1050m)=ANSWER=184 m/s @= (vox)/v[(vox)^2+(voy)^2+2ay(y)]=cos ^-1[(115ms)/(115ms)^2+(0ms)^2+(-9. 8ms^2)-(-1050m)= ANSWER 51. 3 deg A placekicker kicks a football at an angle of __40. 0° above the horizontal axis. The initial speed of the ball is Vo=22 m/s. Ignore air resistance, and find the maximum height H that the ball attains. Y=H=[(Vy^2)-(Voy^2)]/(2ay)= (0ms)^2-(14ms)^2/2(-9. 8ms^2)= ANSWER +10 m
It is also possible to find the total time or “hang time”. For the motion, ignore air resistance and use the data from Example 6 to determine the time of flight between kickoff and landing. (y =Voy(t)+ ½ ay( t^ 2)=(14ms)+ ½ (-9. 8ms^2)t=0 ANSWER t=2. 9s Ignore air resistance and calculate the range R of the projectile. Vox=Vo([email protected])= +(22ms)(cos40 deg)= +17m/s x=R=Vox(t)= +(17ms)(2. 9s)= ANSWER +49 m The engine of a boat drives it across a river that is 1800 m wide. The velocity Vbw of the boat relative to the water is 4. 0 m /s, directed perpendicular to the current. The velocity Vws of the water relative to the shore is 2. m /s. (a) What is the velocity Vbs of the boat relative to the shore? (b) How long does it take for the boat to cross the river? a) Vbs=v(Vbw)^2+(Vws)^2=v(4. 0ms)^2+(2. 0ms)^2= ANSWER 4. 5 m/s b) t= width/(Vbs)[email protected]= (1800 m)/(4. 0 m/s)= ANSWER 450 s Suppose you are driving due east, traveling a distance of 1500 m in 2 minutes. You then turn due north and travel the same distance in the same time. What can be said about the average speeds and the average velocities for the two segments of the trip? Answer (b) The average speeds are the same, but the average velocities are different.
A power boat, starting from rest, maintains a constant acceleration. After a certain time t, its displacement and velocity are r and v. At time 2t, what would be its displacement and velocity, assuming the acceleration remains the same? ANSWER (c) 4r and 2v A projectile is fired into the air, and it follows the parabolic path shown in the drawing, landing on the right. There is no air resistance. At any instant, the projectile has a velocity v and an acceleration a. Which one or more of the drawings could not represent the directions for v and a at any point on the trajectory?
ANSWER A (angle is less that 90deg with vertex pointing to left) and C (right angle) An object is thrown upward at an angle _ above the ground, eventually returning to earth. (a) Is there any place along the trajectory where the velocity and acceleration are perpendicular? If so, where? ANSWER yes where obj is at its highest point (b) Is there any place where the velocity and acceleration are parallel? If so, where? ANSWER No A tennis ball is hit upward into the air and moves along an arc. Neglecting air resistance, where along the arc is the speed of the ball (a) a minimum?
ANSWER when the ball is at its highest point in the trajectory (b) a maximum? ANSWER initial and final position of the motion A rifle, at a height H above the ground, fires a bullet parallel to the ground. At the same instant and at the same height, a second bullet is dropped from rest. In the absence of air resistance, which bullet, if either, strikes the ground first? ANSWER both bullets reach ground at same time A leopard springs upward at a 45 deg angle and then falls back to the ground. Air resistance is negligible. Does the leopard, at any point on its trajectory, ever have a speed that is one-half its initial value?
ANSWER No Two balls are launched upward from the same spot at different angles with respect to the ground. Both balls rise to the same maximum height. Ball A, however, follows a trajectory that has a greater range than that of ball B. Ignoring air resistance; decide which ball, if either, has the greater launch speed. ANSWER ball A has greater launch speed A man is stranded on a raft (mass of man and raft = 1300 kg. By paddling, he causes an average force P of 17 N to be applied to the raft in a direction due east (the + x direction). The wind also exerts a force A on the raft.
This force has a magnitude of 15 N and points 67 deg north of east. Ignoring any resistance from the water, find the x and y components of the raft’s acceleration. X component 15 N x (cos 67)= 6N 17 N + 6 N= 23 N (ax)= E(Fx)/m= (+23N)/(1300kg)= ANSWER +0. 018 m/s^2 Y component 15N x (sin 67)= 14 N 0 N + 14 N= 14 N (ay)= E(Fx)/m= (+14N)/(1300kg) = ANSWER +0. 011 m/s^2 The mass of the Hubble Space Telescope is 11 600 kg. Determine the weight of the telescope (a) when it was resting on the earth W=G(Mg m)/(r^2)= [(6. 67 x 10^-11 N x m^2/kg^2)(5. 98 x 10^24kg)(11600kg)]/(6. 8 x 10^6m)^2= ANSWER 1. 14 x 10^5 N (b) as it is in its orbit 598 km above the earth’s surface. (use same equation used in A but change bottom to 6. 98 x 10^6m) ANSWER 0. 950 x 10^5 N All of the following, except one, cause the acceleration of an object to double. Which one is the exception? (a) All forces acting on the object double. (b) The net force acting on the object doubles. (c) Both the net force acting on the object and the mass of the object double. (d) The net force acting on the object remains the same, while the mass of the object is reduced by a factor of two.
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