# Ap Physics kinematics help!

jai6638
Question 1) Graphically determine the resultant of the following three vector displacements:

a) 34m , 25 degree north of east
b) 48 m, 33 degree east of north
c) 22m, 56 degree west of south.

Ax = 30.814
Ay= 13.369

Bx= 40.256
By= 26.142

Cx=x 12.30
Cy= 18.238

I added the x components and y components and used the pythagorean theorm to get 101 m as my resulant.. HOwever, the book says the answer is 58 m... need some help...

thanks much

Mentor
jai6638 said:
a) 34m , 25 degree north of east
b) 48 m, 33 degree east of north
c) 22m, 56 degree west of south.

Ax = 30.814
Ay= 13.369
Recheck this value.

Bx= 40.256
By= 26.142
x and y are mixed up. (That's 33 degrees east of north.)

Cx=x 12.30
Cy= 18.238
Again, x and y are mixed up. And the sign is wrong.

Pay more attention to the angles. Draw a diagram of each vector.

jai6638
Ax = 30.814
Ay= 14.369

Bx= 26.14 ( since I took theta as 90-33 = 57 degrees )
By= 40.25 ( since I took theta as 90-33 = 57 degrees )

Cx= -18.238 ( ( since I took theta as 90-56 = 34 degrees )
Cy= -12.30 ( ( since I took theta as 90-56 = 34 degrees )

is this correct?

thanks much.

Last edited:
Mentor
Recheck the sign of the C components.

jai6638

Mentor
Seems like you're just guessing. Instead, actually draw an arrow pointing south, then another pointing west of south. What are the signs of the x and y components of that arrow?

(Note: North is +, South is -; East is +, West is -.)

jai6638
well I have a feeling that my diagram is wrong and hence I wasnt referring to it when I edite dthe answers..

My diagram is setup like this:

starting from the same origin, I have three lines: a, b , c ... A goes north of east at 25 degrees. B goes east of north at 33 degrees and C goes 56 degrees west of south ( and hence all the x and y values are negative ) .. Is this correct?

EDIT: just realized while Iwas writing my diagram setup that both components of C will be negative since it is west of south and not west of north which is what i was reading it as..

Mentor
jai6638 said:
EDIT: just realized while Iwas writing my diagram setup that both components of C will be negative since it is west of south and not west of north which is what i was reading it as..
Now you've got it!

jai6638
Even though I did this and got the answer, I had a question.. Is using the components of 90-33= 57 degrees north of east correct? shouldn't it yield a different result than 33 degrees east of north? or is it one and the same thing?

Mentor
57 degrees north of east is the same as 33 degrees east of north. To see this, draw yourself a picture.

What you may not realize is that cos(33) = sin(57), and sin(33) = cos(57). So either way, you'll get the same answer. (As long as you don't make a mistake.)

jai6638
I see .. cool thanks..

Another questoin:

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horiszoontally with speed of 250 km/hr ( 69.4m/s) .

A) how far in advance of the receipients must the goods be dropped?

I found this to be 480 m which is correct.

B) Suppose instead that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical vleocity ( up or down) should the supplies be given so that they arrive precisely at the climbers positiron?

c) With what speed do the supplies land in the latter case?

HOw do i go about solving B first of all? Do i use X=Vox . t and try to find T and then plug it into the y=y0+v0t-.5gt^2 and find y... Then plug in y into y/t= V ?? I tried that but I don't get the right answer ( which textbook says is 8.37 m/s

Thanks much.

Mentor
jai6638 said:
HOw do i go about solving B first of all? Do i use X=Vox . t and try to find T and then plug it into the y=y0+v0t-.5gt^2 and find y... Then plug in y into y/t= V ?? I tried that but I don't get the right answer ( which textbook says is 8.37 m/s
Use $x = v_x t$ to find the time. Then use the vertical motion equation to solve for v0: you already know t, y0, and y.

jai6638
well I get the v0 as 16.20 m/s and not 8.37 s ...

I solved the following equation: 0=235+(6.124)v0 - 4.9 ( 6.124)^2 to get v0 ..

jai6638
nevemrind.. I got 8.37 m/sec...

For C) i used v^2= v0^2-2a(y-y0)
v^2= (8.37)^2 - 19.6( 235)
v= 67 m/sec and not 97m/sec

what am I doing wrong?

I apologize for askin you so many qustions.. got a test tomorrow so am trying to make sure I understand the homework probs..

THanks much.

Mentor
jai6638 said:
For C) i used v^2= v0^2-2a(y-y0)
v^2= (8.37)^2 - 19.6( 235)
v= 67 m/sec and not 97m/sec
y-y0 = -235

jai6638
the answer still remains 68 m/sec though unless I am doing something wrong again.. doh..

jai6638
ahh I see.. there's probably an error in the textbook then...

Once again, thanks much :) .. appreciate your help..

Mentor
It's not an error in the book. 68 m/s is just the vertical component of the velocity. You have to also include the horizontal component to get the total speed.

jai6638
Ahh I see.. alright cool... thanks..