AP Physics

Unit 2 - Vectors

Day #6Relative Velocity and i/j/k notation

We’re here on your schedule

#59 The Long John Silver Solution (the easy way)

A

D

B

C

E

Step #1: Need to go 30 right and 100 up to get from A to B. 1/2 of this is 15 right and 50 up. Since you start at A(30, -20), then you finish at W(45, 30).

Step #2: Need to go 55 left and 40 down to get from W to C. 1/3 of this is 18.33 left and 13.33 down. Since you start at W(45, 30), then you finish at X(26.66, 16.66).

Step #3: Need to go 13.33 right and 46.66 down to get from X to D. 1/4 of this is 3.33 right and 11.66 down. Since you start at X(26.66, 16.66), then you finish at Y(30, 5).

Step #4: Need to go 100 left and 55 up to get from Y to E. 1/5 of this 20 left and 11 up. Since you start at Y(30, 5), then you finish at Z(10, 16).

The Long John Silver Solution (the physics way)

A

D

B

C

E

Step #1: Start at tree A and move toward tree B, covering only 1/2 the distance.

First, find the position of B relative to A.

rBA = rBG + rGA = rBG - rAG = (60 i + 80j) - ( 30 i – 20 j) = 30 i + 100j

(1/2) * rBA = 15 i + 50j

rS1G = position at the end of step #1 = rAG + (15 i + 50 j) = 45 i + 30 j

Step #2: Move toward tree C, covering 1/3 the distance.First, find the position of C relative to rS1G.

rCS1 = rCG + rGS1 = rCG – rS1G = (-10 i - 10 j) - ( 45 i + 30 j) = -55 i – 40 j

(1/3) * rCS1 = -18.33333 i - 13.33333 j

rS2G = position at the end of step #2 = rS1G + (-18.33333 i - 13.33333 j) = 26.6666 i + 16.66666 j

G: the origin

Step #3: Move toward tree D, covering 1/4 the distance.

Step #4: Move toward tree D, covering 1/5 the distance.First, find the position of E relative to rS3G.

rES3 = rEG + rGS3 = rEG – rS3G = (-70 i + 60j) - ( 30 i + 5 j) = -100 i + 55 j

(1/5) * rES3 = -20 i + 11 j

rS2G = position at the end of step #1 = rS1G + (-20 i + 11 j) = 10 i + 16 j

First, find the position of C relative to rS2G.

rDS2 = rDG + rGS2 = rDG – rS2G = (40 i - 30 j) – (26.6666 i + 16.66666 j) = 13.33333 i – 46.66666 j

(1/4) * rDS2 = 3.33333i - 11.66666 j

rS3G = position at the end of step #3 = rS2G + (3.33333i - 11.66666 j) = 30 i + 5 j

Dig here!!!

Problem #61 Solution

+ =

sin = 3/5 = sin-1(3/5)=36.87o (notice that and equals the same value)

These two x-components must cancel each other out

Read & learn the rest of Chapter 3. Solve problems 3, 10, 23, 28, 31, and 33 on pp 65-67.

TONIGHTS HW

X: 50 +32cos40 – 16cos30 = 60.675 km

Y: 0 +32sin40 – 16sin30 = 12.569 km

r = (60.675i + 12.569j) km

r| 61.96 km

Sincer 61.96 km [E 11.7o N], the car must travel _____________________ to get back.

61.96 km [W 11.7o S]

40o

30o

11.7o

€

A + B = (3i − 5 j) + (−7i − 9 j)

€

−4i −14 j

€

A − B = (3i − 5 j) − (−7i − 9 j)

€

10i + 4 j

€

(10.77, 21.8o)

€

(14.59, 74.1o)

€

C = 4i +14 j

€

v1 = (50i)mph

€

v2 = (32sin 45i − 32cos45 j)mph

€

22.627i − 22.627 j

€

a =v2 − v1

Δt

€

(22.627i − 22.627 j)mph − (50i)mph

4sec

€

(−27.373i − 22.627 j)mph

4sec

€

(−6.84i − 5.66 j)mi /h /s

€

(−10.03i − 8.30 j) fts2

€

13.02 fts2 [W 39.6oS]

€

VPG =VPW +VWG

€

200mph[W 20oN] =VPW +10mph[N]

€

VPW = 200mph[W 20oN] −10mph[N]

€

VPW = −200cos20i + 200sin20 j( ) − 10 j( )

€

VPW = −187.94i + 58.40 j

€

196.8mph

€

[W17.3oN]

€

x : 60cos(−60o) + 35cos60o + 200cos(−140o)

€

30 +17.5 −153.2

€

−105.71N

€

y : 60sin(−60o) + 35sin60o + 200sin(−140o)

€

−51.96 + 30.31−128.56

€

−150.21N

€

F4 = (_______ i + _______ j)N

€

105.71

€

150.21

€

183.68N

€

@ 54.86o above the + x - axis

€

|F |= 52 + 32 + 62 = 8.367

€

8.367

€

5

x

€

8.367

€

3

y

€

8.367

€

6

z

€

x = cos−1(5 /8.367)

€

53.31o

€

y = cos−1(3/8.367)

€

68.99o

€

z = cos−1(6 /8.367)

€

44.18o