Physics Vector Problem Need HELP!?

Question

Dennis has a problem. He went to school, which is 12 blocks at 60° away from home. Then he went to the intramural scrimmage at Dumbbell High Football field 14 blocks at 315° from school. Then he had a study date with paula and her house is -12 blacks x from the football field. The library they studied at is 11 blocks y from Paula's. After studying, they went to go see a romantic comedy at the movie theatre, -8.5x - 8.5y from the library. Paula's dad is the protective kind, so her parents picked up Paula up at the theatre. Dennis will run out of gas 5 blocks away from the theatre according to his high tech gas gauge. If that happens, can Dennis make it home on these weird city streets (assume dennis can travel in any direction)? If yes, what direction does Dennis need to travel in from the movies? If no, how close can he get to home?

Answer 1

The way I'd do this problem is:

Make X and Y coordinate columns...

Add up all the coordinates for your travel...

X                                                

12cos(60)                          

14cos(-45)                          

-12                                      

0                                        

-8.5                                    

Y

12sin(60)

12sin(-45)

0

11

-8.5

This adds up to be roughly -4.6x and 3y.

Use the Pythagorean theorem to get the distance you are from your starting point.

D = sqrt(4.6^2 + 3^2) = 5.49

This means you have about a 1/2 block to walk.

Answer 2

In the following, i, j are unit vectors along x,y axes respectively.

All distance measurements are in blocks.

Let AB = displacement from Dennis's home to school

BC = displacement from school to intramural scrimmage at football field

CD = displacement from football field to Paula's house

DE = displacement from Paula's house to library

EF = displacement from library to movie theatre

AB = 12 cos 60°i + 12 sin 60°j = 6 i + 6 √3 j

BC = 14 cos 315° i + 14 sin 315°j = 14 cos (360 - 45)° i + 14 sin (360 - 45)°j

Or, BC = 14 cos(-45°) i + 14 sin (-45°) j

cos(- θ) = cos θ and sin (- θ) = - sin θ

Therefore, BC = 14 cos 45° i - 14 sin 45° j = 14/√2 i - 14/√2 j

Or, BC = 7√2 i - 7√2 j

CD = -12 i

DE = 11 j

EF = - 8.5 i - 8.5 j

Total displacement from Dennis's house to movie theatre = AF = AB + BC + CD + DE + EF

Or, AF = 6 i + 6 √3 j + 7√2 i - 7√2 j - 12 i + 11 j - 8.5 i - 8.5 j

Or, AF = 6 i + 7√2 i - 12 i - 8.5 i + 6 √3 j - 7√2 j + 11 j - 8.5 j

Or, AF = (7√2 - 14.5)i + (6 √3 - 7 √2 + 2.5)j

Or, AF = (7*1.414 - 14.5)i + (6 * 1.732 - 7 * 1.414 + 2.5)j

Or, AF = -4.6 i + 3j

Or, FA = -AF = 4.6 i - 3 j

Shortest distance from thetre to Dennis's house = |FA| = √{4.6^2 + (-3)^2} = √(21.16 + 9) = √(30.16) = 5.5

Dennis runs out of gas 5 blocks away from thetre. This means that he cannot make it home. At best, he can reach upto 0.5 blocks away from home.

For that he has to travel along FA i.e. at angle arctan(-3/4.6) = 327° i.e. 327 * pi/180 = 1.8 π radian from the theatre.

Ans: Dennis has to travel 327° = 1.8 π radian from the theatre. He will not make it to home. He will reach to a distance of half block from home.