Changes in velocity with Direction change

[skatergirl]

Homework Statement

Tim is running cross country at 6.4m/s when he completes a wide angle turn and continues at 5.8m/s[w]. What is his change in velocity?

Homework Equations

Δv=v₂-v₁
a²+b²=c²

The Attempt at a Solution

i am not sure how i am supposed to find the change in velocity...
i could just find the hypotenuse but i think that is giving me the resulting acceleration and not the actual change. but I am quite sure its not as simple as just subtracting the two values either ??

 

[Snark1994]

You have to remember that velocity is a vector: so we have v_1 = (0,6.4) and v_2 = (5.8,0) (where (1,0) points West and (0,1) points South), so as you correctly wrote: \Delta v = v_2 - v_1 = (5.8, -6.4). This is technically the change in his velocity, but the question may be just asking for its magnitude (which, as you note, is the hypotenuse).

 

[skatergirl]

ok i will include both in my answer then. thank you. also should i be including the angle/direction? do you draw the s or w arrow first when drawing the diagram? like will the angle be between the hyp and 5.8 or the hyp and 6.4?

 

[Snark1994]

also should i be including the angle/direction?

If you include the information about what your axes are (i.e. South = (0,1), West = (1,0) ) then when you write the components of \Delta v down you have represented it unambiguously. You shouldn't need to include it unless the question specifically asks you for it.

do you draw the s or w arrow first when drawing the diagram? like will the angle be between the hyp and 5.8 or the hyp and 6.4?

I'm not sure I understand the first question. In answer to the second, you would have to be specific (e.g. "the vector is 10° to the East of North", or "at a bearing of 010°" or some similar specification of angle - and note that was just an example, not the direction of \Delta v)