NO ME SALEN
SOLVED PHYSICS EXCERCISES FROM THE CBC
Cinemática vectorial 


Aditional No me salen 1.10*  An automovil displaces Northward at 50km/h, then, it displaces Eastward, for 4 hours, travelling 400km in the process. Finally, it heads to the South, at 100km/h for 300km. The average speed modulus is:

a) 40 km/h
d) 100 km/h

b) 50 km/h
e) 30 km/h 
c) 0 km/h
f) 25 km/h 



A really simple excercise that had a very low performance, only a few students could solve this correctly in the exam. Most of the calculations and the general procedure is so easy than you can (and should) solve it mentally
I´ll solve it step by step as I would (almost mentally) have done, being I the student at the exam. 

The first displacemen, Δr_{1}, to the north, goes for three hours, at 50 km/h. Then, in this first movement, it travels 150 km.
The second displacement, Δr_{2},was 200km to the East, it lasted for 4 hours.
Finally, the third displacement, Δr_{3}, was 300km to the South, at 100km/h, that means it lasted for 3 hours.
The average total speed, the one we are asked by the excercise, is the quotient between the total displacement, Δr_{T}, and the time spent in that total displacement.
The total displacement is in green, the vector wich has it´s origin in the origin of the first displacement, and it´s head in the last displacement´s head. It´s just the vectorial sum of the vectors. 



To know their modulus, you can apply the Pitagoras Theorem in the yellow colored triangle up there.
Δr_{T}² = (200 km)² + (150 km)²
Δr_{T} = 250 km
the total length is the sum of the lengths of each section. The first was 3 h, the second 4 and the last 3. In total, Δt_{T}, 10 hours. Then:
v_{m} = Δr_{T} / Δt_{T}
v_{m} = 250 km /10 h



v_{m} = 25 km/h 
option f) 


It was not so grievous
CHALLENGE: express the total average speed in vector notation, according to the reference system adopted in the scheme. 




*Excercise taken at the first midterm exam in University city, first quarter 2015. 




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