Displacement, Elapsed Time, Magnitude, Direction, and Average Speed on a Trip.

[CaptFormal]

Homework Statement

A group of physics students take off in a small plane to visit physics Land. The plan flies at a constant speed of 160 mph. They start at Newtonville and travel 160 miles due north to Faraday Town. At this point they change course for Maxwell City, which is 320 miles away at a heading of 20 degrees North of East. At Maxwell City, they take an hour-long break. They depart for Bohr Village, which is 120 miles away, at a heading of 15 degrees East of South. The final leg takes them 80 miles due south to Einstein City.

1. What is their displacement for the trip? Give this in terms of how far North/South and East/West they end up from Newtonville.

2. What was the elapsed time?

3. What is the magnitude and direction of the average velocity?

4. Find the average speed for the entire trip.

Homework Equations

Vx = Vcos(theta)
Vy = Vsin(theta)

Average Speed = (total distance)/(total time)

The Attempt at a Solution

(First Question)
So, I think I went a little too far in my calculations and did not quite understand what the problem was asking me. First I figured out all the (x) and (y) components of each of the velocities between each of the cities which is shown below:

A = From Newtonville to Faraday Town
B = From Faraday Town to Maxwell City
C = From Maxwell City to Bohr Village
D = From Bohr Village to Einstein City

Vax = 160cos(90) = 0
Vay = 160sin(90) = 160

Vbx = 320cos(20) = 300.702
Vby = 320sin(20) = 109.446

Vcx = 120cos(255) = -31.06
Vcy = 120sin(255) = -115.91

Vdx = 80cos(270) = 0
Vdy = 80sin(270) = -80

Then I took all of these individual velocity components and found the overall velocity of the trip by the following:

V = [(vx)^2 + (vy)^2]^(1/2)

V = [(0+31.06+300.702+0)^2 + (160+115.91+109.446-80)^2]^(1/2)

V = 270.94

Now to find the direction I did the following:

tan-1(sum of vy/sum of vx) = degrees

tan-1 (73.536/269.64) = 15.3 degrees

I concluded that they ended up 270.94 miles away from Newtonville and 15.3 degrees from the East. I got this wrong because I only needed components. Does that mean I should have forgone finding the degrees?

(Second Question)

A = Time between Newtonville and Faraday Town
B = Time between Faraday Town and Maxwell City
C = Time between Maxwell City and Bohr Village
D = Time between Bohr Village and Einstein City

So, First I found the amount of time it took to travel to each city:

A = 160 miles*(1 hr/160 miles) = 1 hr

B = 320 miles*(1 hr/160 miles) = 2 hrs

C = 120 miles*(1 hr/160 miles) = 0.75 hr

D = 80 miles*(1 hr/160 miles) = 0.5 hr

Total trip = A + B + C + D

Total trip = 4.25 hrs.

Unfortunately I forgot to add the hour long break that they took at Maxwell City, so the total elapsed time should be 5.25 hours. Am I correct? Is there anything else I am missing which could have this problem be incorrect?

(Question 3)

So, I thought that since they just wanted the magnitude and direction of the average velocity that I already found this out in question 1. So I said that the magnitude was 270.94 at 15 degrees. But, I was told that I found that 270.94 is the displacement, not the average velocity. What must I do in order to find the average velocity?

(Question 4)

So, in order to find the average speed I did the following:

Average Speed = (Total Distance)/(Total Time)

Average Speed = (160 miles + 320 miles + 120 miles + 80 miles)/(4.25 hours) = 160 miles/hr

I was told that I should have included the time it took for the break which is I think the mistake I made for question 2. If I did include the time it took for the break I would get an average speed of 129.524 miles/hour. Is this correct?

I realize this is a long problem and it really was a challenge for me to calculate. Any assistance would be greatly appreciated. Thank you for your time.

 

[anathema]

Dear forum post,

Thank you for sharing your work and questions regarding this physics problem. I am a scientist and I would like to help you with your calculations and understanding of the problem.

1. To find the displacement, we need to find the final position of the plane relative to Newtonville. We can do this by adding up all the x and y components of the velocities.

Vx = Vcos(theta)
Vy = Vsin(theta)

For the x-direction, we have:
Vx = 160cos(90) + 320cos(20) + 120cos(255) + 80cos(270)
Vx = 0 + 300.702 + (-31.06) + 0 = 269.642 miles

For the y-direction, we have:
Vy = 160sin(90) + 320sin(20) + 120sin(255) + 80sin(270)
Vy = 160 + 109.446 + (-115.91) + (-80) = 73.536 miles

Therefore, the displacement is 269.642 miles east and 73.536 miles north of Newtonville.

2. You are correct in your calculation of the elapsed time. We need to add the one-hour break at Maxwell City to get a total of 5.25 hours.

3. The average velocity is the displacement divided by the elapsed time. We have already calculated the displacement in question 1 and the elapsed time in question 2. Therefore, the average velocity is:

Vavg = (269.642 miles, 73.536 miles) / 5.25 hours = (51.42 miles/hour, 14.01 miles/hour)

The magnitude of the average velocity is given by the Pythagorean theorem:

|Vavg| = [(51.42 miles/hour)^2 + (14.01 miles/hour)^2]^(1/2) = 52.95 miles/hour

The direction of the average velocity can be found using trigonometry:

tan(theta) = (73.536 miles) / (269.642 miles)
theta = tan^-1(73.536 miles / 269.642 miles) = 15.3 degrees east of north

4. To find the average speed, we need to divide the total distance traveled by the total time, including the one-hour break at Maxwell City.

Average speed = (160 miles