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PHY 231 1 PHYSICS 231 Lecture 4: Vectors Remco Zegers Walk-in hour: Thu. 11:30-13:30 Helproom.

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Presentation on theme: "PHY 231 1 PHYSICS 231 Lecture 4: Vectors Remco Zegers Walk-in hour: Thu. 11:30-13:30 Helproom."— Presentation transcript:

1 PHY 231 1 PHYSICS 231 Lecture 4: Vectors Remco Zegers Walk-in hour: Thu. 11:30-13:30 Helproom

2 PHY 231 2 Vectors and Scalars Scalar: A quantity specified by its magnitude only Vector: A quantity specified both by its magnitude and direction. To distinguish a vector from a scalar quantity, it is usually written with an arrow above it, or in bold to distinguish it from a scalar. Scalar: A Vector: A or A

3 PHY 231 3 Question Are these two vectors the same? Are the lengths of these two vectors the same? Two vectors are equal if both their length and direction are the same!

4 PHY 231 4 Vector addition A B A+B A B B+A A+B=B+A

5 PHY 231 5 Vector subtraction A -B A-B=A+(-B) B

6 PHY 231 6 Vector operations in equations A B A+B y x (x a,y a ) (x b,y b ) (x a+b,y a+b ) Example:

7 PHY 231 7 Question begin end Which route is shorter?

8 PHY 231 8 The length of a vector and its components Y x (x a,y a ) Length of vector (use pythagorean theorem): 

9 PHY 231 9 Question A man walks 5 km/h. He travels 12 minutes to the east, 30 minutes to the south-east and 36 minutes to the north. A) What is the displacement of the man? B) What is the total distance he walked? 1 km 2.5 km 3 km x 2 =2.5cos(  )=1.77  =315 o Y 2 =2.5sin(  )=-1.77 A) B) 1+2.5+3=6.5 km

10 PHY 231 10 Relative motion Motion is relative to a frame! A woman in a train moving 50 m/s throws a ball straight up with a velocity of 5 m/s. A second person watches the train pass by and sees the woman through a window. What is the motion of the ball seen from the point of view from the man outside the train? Motion of the ball in rest-frame of train Motion of the train Resulting motion

11 PHY 231 11 Question  A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. 1) At what angle  does the captain have to steer the boat the go straight across? A) 30 o B) 45 o C) 0 o D) -45 o 2) how long does it take for the boat to cross the river? A) 6 min B) 6.9 min C) 12 min D) 1 h 3) If it doesn’t matter at what point the boat reaches the other side, at what angle should the captain steer to cross in the fastest way? A) 30 o B) 45 o C) 0 o D) -45 o

12 PHY 231 12 Answer  Counter balance flow=5km/h Flow=5km/h Maximum v=10 km/h Hyp. Opp. 1) sin  =opposite/hypothenuse =5/10=0.5  =sin -1 0.5=30 o adjacent 2)tan  =opposite/adjacent tan30 o =0.577=5/velocity hor velocity hor =8.66 km/h time=(1 km)/(8.66 km/h)= 0.115 h=6.9 min 3) 0 o (the horizontal component of the velocity is then maximum.


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