Physics 1307 Quiz Chapters 23 Solution 1/2 NW θ ABD
Physics 1307 Quiz Chapters 23 Solution
Problem 1 The eye of a hurricane passes over Grand Bahama Island. It is moving in a direction 60.0º north of west with a speed of 41.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h. How far from Grand Bahama is the hurricane 4.50 h after
This problem combines vector addition with motion with uniform velocity. Let us make first a graphic of the situation described by the problem.
D
B
N
A W
The first displacement is given by : t ={−41.0km/ hcos600 ,41.0km/hsin600 }3 h=−61.5,106.5km A=V A while the second displacement is equal to : t={0, 25.0km/ h}1.5h=0,37.5km B=V B Where we have used the fact that the second interval is equal to 4.5 h 3.0 h. Now, the total displacement is just the sum of the two previous ones thus D= A B=−61.5 ,144km . Its magnitude gives the distance from Gran Bahama −61.5 21442=156.6km ∣D∣=
1/2
Problem 2 A racing car reaches a speed of 40 m/s. At this instant, it begins a uniform negative acceleration, using a parachute and a braking system, and comes to rest 5.0 s later. (a) Determine the acceleration of the car. (b) How far does the car travel after acceleration starts? v i = 40 m/s
vf= 0
(a) Since we are given the time elapsed and the initial and final velocities we can use the definition of acceleration to find its value: a=
vf − vi 0−40 m/ s 2 = =−8m/ s t 5.0s
(b) The distance after the deceleration stars is : 1 x=v ox t at2 2 1 x=40m/ s 5.0s −8m/s 5.0s2 2 x=100m
2/2
Problem 1 The eye of a hurricane passes over Grand Bahama Island. It is moving in a direction 60.0º north of west with a speed of 41.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h. How far from Grand Bahama is the hurricane 4.50 h after
This problem combines vector addition with motion with uniform velocity. Let us make first a graphic of the situation described by the problem.
D
B
N
A W
The first displacement is given by : t ={−41.0km/ hcos600 ,41.0km/hsin600 }3 h=−61.5,106.5km A=V A while the second displacement is equal to : t={0, 25.0km/ h}1.5h=0,37.5km B=V B Where we have used the fact that the second interval is equal to 4.5 h 3.0 h. Now, the total displacement is just the sum of the two previous ones thus D= A B=−61.5 ,144km . Its magnitude gives the distance from Gran Bahama −61.5 21442=156.6km ∣D∣=
1/2
Problem 2 A racing car reaches a speed of 40 m/s. At this instant, it begins a uniform negative acceleration, using a parachute and a braking system, and comes to rest 5.0 s later. (a) Determine the acceleration of the car. (b) How far does the car travel after acceleration starts? v i = 40 m/s
vf= 0
(a) Since we are given the time elapsed and the initial and final velocities we can use the definition of acceleration to find its value: a=
vf − vi 0−40 m/ s 2 = =−8m/ s t 5.0s
(b) The distance after the deceleration stars is : 1 x=v ox t at2 2 1 x=40m/ s 5.0s −8m/s 5.0s2 2 x=100m
2/2
