HW2: Vectors and 2D Motion

1

5. b_x = -bsinβ

Explanation:

the x component of vector b= b_x

sinθ = o/h = -b_x/b

b_x is in pointing toward the negative x direction

Why is vector b +, magnitude is never -

b_x component shows that vector b heads in the - x direction

θ = β
b_x = -bsinβ

2

1. b_y = bcosβ

Explanation:

the y component of vector b= b_y

cosθ = a/h = +b_y/b

b_y is in pointing toward the + y direction

θ = β
b_y = bcosβ

3-5

3) -114.95 km
4) 111.35 km
5) 45.91º

Explanations:

3) r_x = ?
vector r = sum of vectors a, b, & c
r_x = sum of x components of a, b, & c
a_x = acosα
b_x = -bsinβ
c_x = 154 km
r_x = a_x + b_x + c_{x =} acosα + -bsinβ +154 km = -114.95 km

4) r_y = ?
r_y = sum of y components of a, b, & c
a_y = asinα
b_y = bcosβ
c_y = 0
r_y = a_y + b_y + c_{y =} asinα + bcosβ = 111.35 km

5) find θ West of North of vector r
given: θ = γ, r_x = -114.95 km, r_y = 111.35 km
unknown: vector r
tan θ = opposite/adjacent = r_x/r_y
tan^-1(-114.95/111.35) = 45.91º

726.68 m

Explanation:

given: V₁ = 660 m/h, θ₁ = 47º, V₂ = 600 m/h, θ₂ = 88º, t = 1.7 h

1) find distance traveled by A1 & A2
both planes start at origin
v = d/t → d = v*t
d₁ = 660×1.7h = 1122 m
d₂ = 600×1.7h = 1020 m

2) find x components of d₁ & d₂ using to θ_{1 &} θ₂ to find position coordinates
d_1x = d₁cosθ₁ = 740.52
d_1y = d₁sinθ₁ = 842.91
d_2x = d₂cosθ₂ = 35.597
d_2y = d₂sinθ₂ = 1019.38
A1 is at (740.52, 842.91)
A2 is at (35.597, 1019.38)

3) use distance formula to find distance
d = [(x₂-x₁)² + (y₂-y₁)² ]^1/2 = 726.68 m

7-8

7) 4. i(2 m/s) - j(6 m/s)

8) 3. 0

Explanations:
given: const a = (4 m/s²)j = 4j = 0i + 4j,
at t =2, v = ( i + j) (2 m/s)= 2i + 2j
at t =2, r = (i - j)(4m)

7) find velocity at t = 0 → v(0) = v_i = ?
v_f = 2i + 2j , t = 2
find v_o at t=0 using v_f = v_o + at
2i + 2j = v_o + (0i + 4j)t → v_o = (2i + 2j) - (0×2i + 8j) = (2i-0i) + (2j - 8j) = (2i - 6j)

8) find position at t = 0 → r(0) = r_i = ?
v_f = 2i + 2j , t = 2
r_f = 4i - 4j, t = 2
v_o = 2i - 6j, t = 0
find r_i at t=0 using r_f = r_i + v_ot + ½at²
4i - 4j = r_i + (2i - 6j)t + ½(0i + 4j)t² → 4i - 4j = r_i + (4i - 12j) + (8j)
r_{i =} (4i - 12j) + (8j) - (4i - 4j) = (4+0-4i) + (-12 + 8 - - 4j) = 0

9

t = 0.5379 s

Explanation:
Looking for the time at which the particle will be traveling at 41º with respect to the horizontal.
Diagram:

1. find v_fy and v_fx equations
   v_fy and v_fx are the velocities at when the particle will be traveling at 41º with respect to the horizontal.
   y given: v_oy = 6.3 m/s, a_y = -9.8 m/s²
   x given: v_ox = 0 m/s, a_x = 2.2 m/s²
   y: v_fy = v_oy + a_yt = 6.3 + -9.8t
   x: v_fx = v_ox + a_xt = 0 + 2.2t

2. find t @ θ = 41º
   tanθ = opposite/adjacent = v_fy/v_fx = (6.3 + -9.8t)/(2.2t)
   tan(41º) = (6.3 + -9.8t)/(2.2t) = 0.5379 s

10-11

10) 8.49 m/s

11) 52.088º

Explanation:

10) Find v_fy when the particle is at x_f = 18.4 m
givens: Δx = 18.4 m, const v_x = 6.7 m/s, a_y = 1.9 m/s², v_oy = 0
solving for v_fy because v_xf = 6.7 m/s (const v)
1. find t it takes for the particle to travel to x_f = 18.4 m
solving for time using v_x because t is same in the x and y direction.
v = d/t → t = d/v = 18.4/6.7 = 2.746 s
2. find v_fy
Δx = ½ (v_o + v_f)t CANNOT USE THIS ONE
v_fy = v_oy + a_yt = 0 + (1.9×2.746) = 5.218 m/s
3. find v_f using v_fx and v_fy components
v_f = (v_fx² + v_fy²)^1/2 = 8.49 m/s

11) Find the angle with the wall that the particle strikes
tan^-1(v_fx/v_fy) = tan^-1(6.7/5.218) = 52.088º

12

4

Explanation:

• when the projectile is at the peak/highest point, the v_y will always be 0. 2 & 5 are incorrect.

• when the projectile is in the air at any point there is always acceleration of gravity a_y = -9.8 m/s². 3 is wrong

• 1 & 4 are left. v_x will always be ≠ 0 or else the object is not in motion.

13-15

13) 122.655 m/s

14) 265.15 m

15) 14.71 s

Explanation:

memorize this equation: d = (v_o² sin2θ)/g

13) solve for magnitude of v_o
given: d or Δx = 1460 m, θ = 36º,
d = (v_o² sin2θ)/g → 1460 = (v_o² sin(2×36))/9.8 m/s² (direction doesnt matter just magnitude)
v_o = 122.655 m/s

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14) find Δy at the highest point
cut the projectile motion in half
at the highest point v_yf = 0 m/s
givens: v_fy = 0 m/s, v_oy = V_osin36, a_y = -9.8 m/s²
v_fy² = v_oy² + 2aΔy
Δy = v_fy² - v_oy² / 2a = 265.15m

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15) find t for projectile to reach target
- split the trajectory in half. solve for t_a & x 2
givens: v_o = 122.655 m/s, v_oy = V_osin36, @ t_A v_fy = 0, Δy = 265.15 m, a_y = - 9.8 m/s²
Δy = ½ (v_oy + v_f)t_A → t_A = (2 * Δy)/(v_oy + v_f) = (530.3)/(72.09-0) = 7.3555
t_A * 2 = 14.71 s

16

How high is the fence?

376.09 ft

Explanation:

givens: a_y = -32 ft/s², @ t = 4.9 s v_y = 0, @ t_tot ball hits fence

1) find v_oy at t = 0 using v_fy at peak t = 4.9s
v_fy = v_oy + a_yt → v_{oy =} v_fy - a_yt = 0 - (-32×4.9)= 156.8 ft/s

2) calculate the total time for the ball to hit the fence
t_tot = 4.9 + .71 = 5.61 s

3) calculate Δy
Δy = v_oyt + ½a_yt² = 156.8(5.61) + ½(-32)(5.61)² = 376.09 ft

17

Δy = 33.46 m

Explanation:

1. use equation d = v_o² sin2θ/g to find v_ox
   d = v_o² sin2θ/g = 66.9185 = v_o² sin(2×90)/9.8
   v_o² = 655.80 m/s

2. find Δy of ball thrown up
   v_fy² = v_oy² + 2aΔy
   (0)² = (655.80) + 2(-9.8)Δy
   Δy = 33.46 m

18

MONKEY AND BULLET

Explanation:
the bullet will ALWAYS hit the monkey

1. set up vertical position equations for bullet and monkey
   bullet: y_fb = y_ib+ v_obt + ½a_yt² =
   monkey: y_fm = y_im + v_omt + ½a_yt²

2. set equations equal to solve for t_collide

3. plug t_collide back into any vertical position equation in step 1 to solve for h_f

19. Relative velocity

3. v_b - v_t

Explanation:

1. set up relative velocity equation
   v_bm = v_bp + v_pm
   v_bm /v_b= velocity of the boy relative to the man total because the man is stationary and sees the boy AND train/passengers moving
   v_bp/v_t = velocity of the boy relative to the train/passengers
   v_pm = velocity of train/passengers relative to the man
   v_bm = v_bp + v_pm → v_b = v_t + v_pm

2. solve for velocity of the train with respect to the man (v_pm)
   v_pm = v_b - v_t

20

173.51 m

Explanation:

• asking to find Δx vertical displacement

1. set up relative motion equation
   v_bs = v_bw + v_ws

2. find x and y components of v_bw & v_ws

3. find v_bsx & v_bsy
   add v_bwx & v_wsx, add v_bwy & v_wsy

4. find t it takes to travel the Δy
   v = d/t → t = =d/v

5. use t to find Δx

21

this q lowk rlly hard

22

23

use centripetal acceleration equation

24

Explanation:

when the string breaks a_c = 0 because there is no tension force creating acceleration

1. find v_fy when the ball lands
   givens: v_iy = 0, a_y = 9.8, Δy = -1.53 m
   v² = v^2_o + 2aΔy

2. find the time it takes for the ball to fall
   v_fy = v_iy + gt

3. find v_ox when the ball dettaches and falls

4. find centripetal acceleration using v_ox