Solving 2D Vector Problem - Displacement Calculation

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The discussion revolves around calculating the total displacement of a person who drives 26 km north and then 62 km at an angle of 30 degrees north of east. Participants suggest using the cosine law to solve the problem since it involves two sides and the angle between them. One contributor emphasizes the need for the cosine law to accurately determine the displacement, while another acknowledges a previous oversight regarding the distance traveled on the second highway. The conversation highlights the importance of understanding vector addition in non-right triangle scenarios. The final calculation will yield the total displacement based on these parameters.
jen333
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Hey, i have a 2D vectors problem here:

A person leaves home, drives 26km due north, then turns onto a second highway and continues in a direction of 30.0 degrees north of east. What is her total displacement?

i've drawn a diagram, and the person's 'path' obviously does not make a right angle triangle. therefore I'm not sure how to calculate displacement for this problem. any hints or pointers greatly appreciated!

-jen
 
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jen333 said:
Hey, i have a 2D vectors problem here:

A person leaves home, drives 26km due north, then turns onto a second highway and continues in a direction of 30.0 degrees north of east. What is her total displacement?

i've drawn a diagram, and the person's 'path' obviously does not make a right angle triangle. therefore I'm not sure how to calculate displacement for this problem. any hints or pointers greatly appreciated!

-jen
From what I understand, either you need to know another length (the one of the second highway) or you need to express the displacement in terms of the length of the second highway e.g. you do not have a defined number.

The Bob (2004 ©)
 
oh, sorry! my mistake...
i forgot to add that the person drives for a distance of 62km on the second highway.

-jen
 
Cosine law...
 
While bross7's method would work, I'm almost sure you're supposed to use the cosine law, as Werg22 said.

A hint: you have side a and you have side b. You also have the angle between them (Angle C).
 
You would be quite right. I got lazy and while technically my method would work, it isn't the best fundamentally.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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