I need someone to review the following questions for my summer assignment and tell me if my answers are right or wrong. Please do it carefully, and the problems I found to be the most trickiest were #2 and #5. (This is is AP physics)

Question

#1
In a pickup football game, a receiver stands right next to the quarterback when the ball is hiked. The receiver then runs 10 m north, before veering off at a 30° angle to the left (30° W of N) and running an additional 7 m before they catch the football. What was the receiver’s total displacement from the quarterback (and the football’s, too) when the ball was caught?

A and B are the two vecors I name for each question
<10 cos 90, 10 sin 90> = A
<7 cos 120, 7 sin 120> = B
Resultant: <-3.5, 16.062>
then I squared ^^ both numbers and added them together and finally did square root
Final answer is 16.439 m.

#2
A spy stands on the sidewalk by an apartment building examining a foreign embassy across the street. The spy then climbs 30 m to the top the building and then slides 40 m along a wire at a 20° angle below right to reach a window on the embassy. What is their total displacement from the sidewalk to the embassy window? (I'm not too sure which way the 20 degree wire is going)

The x- components : 40 cos 20= 37.588
The y- components : 30+40 sin 20= 43.681
Resultant: <37.588, 43.681>
Final answer is 57.627 m

#3
An army tank division leaves base and travels 30 miles at 30° S of W and then turns and travels 70 miles at 10° N of W. What is their total displacement from base at the end of the trip?

A= <30 cos 210, 30 sin 210>
B= <70 cos 170, 70 sin 170>
Resultant: <-94.917, -2.845>
Final answer is 94.959 m

#4
A spider starts walking from the center of a piece of paper and walks 10 cm at 15° N of W and then walks 23 cm at 70° S of E before resting. What is the spider’s total displacement from the center to their resting place?
A= <10 cos 165, 10 sin 165>
B = <23 cos 290, 23 sin 290>
Resultant: <-1.793, -19.025>
Final answer is 19.109 cm

#5
A hiker starts at the base of the trail and hikes 350 m up a slope at a 10° angle above right. They come to a switchback and head 500 m up a slope at 7° above left. What is the hiker’s total displacement from the base of the trail when they reach the end of the second leg?

A= <350 cos 10, 350 sin 10>
B= <500 cos 173, 500 sin 173>
Resultant: <-151.590, 121.712>
Final answer is 194.405 m

Answer 1

#1
Using the law of cosines, the distance x is
x^2 = 10^2 + 7^2 - 2*10*7 cos 150°
= 100+49+140*√3/2
= 270.24
x = 16.439
Looks good

#2 is obviously wrong. The street is less than 40m across, and the window is less than 30m high. In that case the distance to the window would be 50m. Your answer is too big.

I think you want y=30+40 sin 20, since he slides down the wire, not up.

Gotta go. Maybe someone else can check 3-5.

Answer 2

@Steve, I already wrote y=30+40 sin 20?

If no one else checks do u mind checking them when you time then?

Answer 3

D = 10m[90o] + 7m[120o].

X = 10*cos90 + 7*cos120 = -3.50 m.
Y = 10*sin90 + 7*sin120 = 16.1 m.

tanAr = Y/X = 16.1/-3.5 = -4.58857
Ar = -77.7o = Reference angle.
A = -77.7 + 180 = 102.3o

D = X/cosA = -3.50/cos102.3 =
16.43 m.[102.3o]

Answer 4

Oops. I meant y=30-40 sin 20

drat that copy/paste!

Answer 5

ahh, so i just have to change the y component to 30-40 sin 20 , and then calculate the displacement again and then its correct?

anyone please check #'s 3,4,5?