Take the vector from O to B and subtract the vector from O to A (which is the same as adding the vector from A to O). The result will be the displacement vector from A to B. The vector components of the vector from A to b will tell you the direction.
Divide the displacement vector magnitude (which will be in km) by 5/6 hr to get the average velocity in km/hr
One last physics problem left:
Peggy drives from City A to City B in 50 minutes. City A is 75.1km from City O in a direction of 25degrees west of south. City B is 23.9km from City O in a direction of 20degrees south of west. Using City O as your origin, find (a) the displacement and (b)Peggy's average velocity for the trip.
answer: Peggy drove____km at___degrees in which direction ,at an average of ____km/hr
8 answers
I got 99, round to 100 for answer as the total displacement at a direction east of south. how do I find the degrees? And for the average velocity I would take 100 divided by 5/6?
You did not do the vector subtraction correctly. Both A and B are in the third quadrant in relation to O.
The components of OA are
-75.1 sin 25 i - 75.1 cos 25 j
= -31.7 i - 68.1 j
The components of OB are
-23.9 cos 20 i - 23.9 sin 20 j
= -22.5 i - 8.17 j.
(The j component is north and the i component is east in my notation)
The vector from A to B is
OB - OA = 9.2 i + 44.2 j
The total displacement is 45.1 km. The direction travelled is east of north by an angle arctan 9.2/44.2 = 11.7 degrees
The components of OA are
-75.1 sin 25 i - 75.1 cos 25 j
= -31.7 i - 68.1 j
The components of OB are
-23.9 cos 20 i - 23.9 sin 20 j
= -22.5 i - 8.17 j.
(The j component is north and the i component is east in my notation)
The vector from A to B is
OB - OA = 9.2 i + 44.2 j
The total displacement is 45.1 km. The direction travelled is east of north by an angle arctan 9.2/44.2 = 11.7 degrees
Ok, so the 45.1 is total displacement, to get the average km/hour do I take the total displacement (45.1) divided by 50 minutes(total time from A to B)...getting 0.902 km/hour, that doesn't seem correct.
that didn't work...this is what the homework website told me was the correct way:
sqrt(23.9cos200degr - )75.1cos245)^2 + (23.9sin200 - 75.1sin 245)^2 = 60.6 km
displacement is
tan-1 {(23.9sin 200)-(75.1sin 245)} / {(23.9cos200)- (75.1cos245)} = 81 degrees north of east
then...60.6 at 81degrees/(50 min)(1hr/60min) = 73 km/h at 81 degrees N of E
I don't understand why they used 200 and 245 instead of the given degrees?
sqrt(23.9cos200degr - )75.1cos245)^2 + (23.9sin200 - 75.1sin 245)^2 = 60.6 km
displacement is
tan-1 {(23.9sin 200)-(75.1sin 245)} / {(23.9cos200)- (75.1cos245)} = 81 degrees north of east
then...60.6 at 81degrees/(50 min)(1hr/60min) = 73 km/h at 81 degrees N of E
I don't understand why they used 200 and 245 instead of the given degrees?
To get km/hr, you divide by 5/6 hour, not 50 minutes.
I may have made a math error in the vector subtraction. The method was correct. That is as far as I have time to go with this.
As far as the angles go, 20 degrees west of south is the same as 200 degrees counterclockwise from east, and 25 degrees west of south is 245 counterclockwise from east. Sines (for north) and cosines (for east) can be used with those angles, and I assume the homework website got it right.
I may have made a math error in the vector subtraction. The method was correct. That is as far as I have time to go with this.
As far as the angles go, 20 degrees west of south is the same as 200 degrees counterclockwise from east, and 25 degrees west of south is 245 counterclockwise from east. Sines (for north) and cosines (for east) can be used with those angles, and I assume the homework website got it right.
oh, I understand now why they used those numbers. thank you for all your input on this physics problem, i really appreciate it.
use the component method to solve the the angle..
1 answer