physics help..?

Question

hey guys, i am having a really tough time trying to answer two physics problems. I am not exactly sure how to go about approaching the two problems, so any help would be greatly appreciated! =)

1) A train moving at a constant speed of 52.0 km/h moves east for 40.0 min, then in a direction 51.0° east of due north for 15.0 min, and then west for 53.0 min. What is the average velocity of the train during this run? (Find magnitude (in km/h) and direction (in degrees)).

2) A plane flies 470 km east from city A to city B in 45.0 min and then 980 km south from city B to city C in 1.50 h. (Assume i points east and j points north.) In magnitude-angle notation, what is the plane's average velocity for the total trip? What is the plane's average speed for the trip?


Thank you so much!

Answer 1

These are vector addition problems:



1) Train motion: Calculate the total displacement, then divide by the time to find the average velocity.

x is the W-E axis, y is the S-N axis.

a) v = 52 km/hr along x; displacement after 40 minutes is therefore (40/60) 52 = 34.666 km



b) v = 52 km/hr at angle 90 - 51 = 39 degrees from x-axis.

Therefore,

v-x = 52 cos(39 deg) = 52 * 0.777 = 40.41



v-y = 52 sin(39 deg) = 52 * 0.629 = 32.72



Since this was for 1/4 of an hour, the displacement during b) is:

X-b = 10.1 km

Y-b = 8.2 km

in 15 minutes.



c) v-x = -52 for 53 minutes

X-c = -52*(53/60) = - 45.9333



So, the total X displacement is:

X = 34.666 +10.1 -45.9333 = -1.1673 km

Y = 8.2 km

and the time was 108 minutes, so this gives

V-x = -1.1673(60/108) = -0.6485 km/hr

V-y = 8.2(60/108) = 4.555 km/hr



The magnitude is sqrt(V-x^2 + V-y^2) = 4.601 km/hr

the direction is Arctan(V-x/V-y) = 8.1 degrees East of North.



2) Plane flight:

a) I'm feeling sleepy, so I won't finish this - I'll just sketch it out:

To calculate the average velocity, calculate the total displacement (as above) and divide by the total time. The first leg is along x, the second along -y. Converting the components into angles you use Arctan as I did above.



b) To calculate average "speed", you divide the total distance (not displacement) by the total time. The distance is 470 + 980, the total time is still 5.25 hours.

Answer 2

Well... For both problems you have to understand the difference between speed and velocity, and between covered distance and displacement :)



For example, if I walk for 3 meters, then turn left and walk for 4 meters, I walked a total of 7 meters. On the other hand, I am only 5 meters away from the point I started from. I have covered a distance of 7 meters, but I moved only for 5 meters ("displacement").



The distance covered is the sum of all the segments (3+4=7); The displacement is the length of the line between the start and end points; in this case sqrt(3^2+4^2)=5.

The displacement can be expressed as a vector stretching from the start point to the end point; for example |r| = 5m; angle: tan(alpha) = 4/3;



Same thing with speed and velocity. Average speed is the distance covered divided by the time elapsed. For example, if I need 10 seconds to move in the manner stated before, I would have an average speed of 7m/10s = 0.7 m/s; It's what a speed meter on an automobile would show (in average).



Velocity is a vector (just like displacement). It has the same direction with displacement, and the value of the displacement divided by elapsed time. In the example above, tan(alpha_velocity) = y/x = 4/3; v = r/t = 5m/10s = 0.5 m/s;



So, for both your problems, all you have to do is get the covered distances, then get the final point in relative coordinates with respect to the start point. Have fun :)



Hope I was of help. Feel free to ask me if there is something you didn't get.

Answer 3

1) Draw 3 vectors. The first one goes directly to the east. Its magnitude is the distance travelled which is speed * time. The second one is 51 degrees east of north, figure out magnitude the same way. Third one is west, again get the magnitude.



Once you've drawn the vectors - and they don't have to be drawn perfectly, just labelled correctly - use trigonometry to figure out the final average direction and magnitude. Distance over total time gives you the final speed.



2) Draw the 2 vectors and make a right triangle. The angle of the hypotenuse is the angle you're looking for, and the length divided by time is the speed.