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 $\Delta v$)

 

[Nickie]

I would approach this problem by first understanding the concept of velocity and how it is affected by changes in direction. Velocity is a vector quantity, meaning it has both magnitude and direction. In this scenario, the initial velocity of Tim is 6.4m/s in the direction of South (represented by ). After completing a wide angle turn, his velocity changes to 5.8m/s in the direction of West (represented by [W]).

To find the change in velocity, we can use the equation Δv=v2-v1, where v2 is the final velocity and v1 is the initial velocity. In this case, v1=6.4m/s and v2=5.8m/s[W]. We can represent these velocities as vectors and use vector subtraction to find the change in velocity.

Using the Pythagorean theorem (a^2+b^2=c^2), we can find the magnitude of the change in velocity by taking the square root of the sum of the squares of the two components (South and West). This gives us a value of 0.9m/s.

Therefore, the change in velocity of Tim after completing the wide angle turn is 0.9m/s, with a direction of South-West. This represents a change in both magnitude and direction of his velocity.