Help with Velocity: "I Thought I Figured It Out

[klm]

i thought i figured it out, but i didnt

A rocket-powered hockey puck starts from the origin and moves on a horizontal frictionless table. In which direction is the puck moving at t=2s? (B) how far from the origin is the puck at t=5for the first part i just thought i needed to look at t=2 s on both graphs, i guessed that vx=15 and vy=30 then i just found the inverse tan to find the direction to be 63.4 degrees. but that is incorrect

 

[PhanthomJay]

i thought i figured it out, but i didnt

A rocket-powered hockey puck starts from the origin and moves on a horizontal frictionless table. In which direction is the puck moving at t=2s? (B) how far from the origin is the puck at t=5

for the first part i just thought i needed to look at t=2 s on both graphs, i guessed that vx=15 and vy=30 then i just found the inverse tan to find the direction to be 63.4 degrees. but that is incorrect

If you look at the slope of the first graph, by determining the vx/t ratio at 5 seconds, you can get exactly what vx is at 2 seconds.

 

[klm]

i'm sorry i do not understand what you mean? because i think that the slope is equal to (40-30)/(5-4) = 10 , but vx does not equal 10 at t=2s

 

[klm]

bump please. i am online so if anyone is willing to help and talk me through it that would be great

 

[learningphysics]

i'm sorry i do not understand what you mean? because i think that the slope is equal to (40-30)/(5-4) = 10 , but vx does not equal 10 at t=2s

slope is (40-0)/(5-0) = 8.

So what is vy at t = 2?

 

[klm]

vy=30 ?

 

[klm]

so does that mean vx=16 and vy=30. so you just do inverse tan (30/16)= 61.9 and that is the direction?

 

[klm]

bump. please can someone tell me if i am thinking about this correctly?

 

[learningphysics]

bump. please can someone tell me if i am thinking about this correctly?

Yes, that's correct.

 

[klm]

thank you very much. can you please help me out with the 2nd part as well? how far from the origin is the puck at t=5

 

[learningphysics]

thank you very much. can you please help me out with the 2nd part as well? how far from the origin is the puck at t=5

area under the v-t graph gives displacement... area under the vx-t graph gives horizontal displacement. area under the vy-t graph gives vertical displacement.

 

[klm]

okay so do you do for vx .5(5x40)=100 and vy (5x30)=150
then do you take the magnitude?

 

[learningphysics]

okay so do you do for vx .5(5x40)=100 and vy (5x30)=150
then do you take the magnitude?

yes, exactly.

 

[klm]

so square root ( 100^2 + 150^2) = 180.28 m ?

 

[learningphysics]

so square root ( 100^2 + 150^2) = 180.28 m ?

yes.

 

[klm]

^when i try to put that in, it says it is incorrect.

 

[learningphysics]

so square root ( 100^2 + 150^2) = 180.28 m ?

oops sorry... it's 180.28cm = 1.8028m

 

[klm]

oohhh thank you so much! i wouldn't have caught that! thank you for your help!

 

[learningphysics]

oohhh thank you so much! i wouldn't have caught that! thank you for your help!

no prob.