# Ap Physics kinematics help!

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

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Doc Al
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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.

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:
Doc Al
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Recheck the sign of the C components.

Doc Al
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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 -.)

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: jsut 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..

Doc Al
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jai6638 said:
EDIT: jsut 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!

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? shouldnt it yield a different result than 33 degrees east of north? or is it one and the same thing?

Doc Al
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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.)

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 dont get the right answer ( which text book says is 8.37 m/s

Thanks much.

Doc Al
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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 dont get the right answer ( which text book 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.

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 ..

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.

Doc Al
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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

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

ahh I see.. there's probably an error in the text book then...

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

Doc Al
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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.

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