HW1: 1D Motion

3. 0 m

displacement = xf- xi. the book begins and ends in the same spot so displacement = 0

3. 8 m

distance traveled = 3m + 3m + 1m + 1m

23.25 m/s
Vo = 15 m/s, constant a= 1.1 m/s², the car accelerates for t = 15 s
The question asks for average speed. To find average speed we will use the equation (Vf-Vo)/2.

Step 1: find Vf
- Using the equation under Motion constant acceleration: vf = vo + at = 15 + (1.1×15) = 31.5 m/s

Step 2: find average speed

• (Vf-Vo)/2 = (31.5-15)/2 = 23.25 m/s

93.001 km

Explanation:

1. Find the distance traveled for d₁, d_2, d_{3
   divide t by 60 min → hr}
   v = d/t → d = t v
   t₁= 28.1 min v₁=70.8 km/h
   t₂= 12 min v₂= 110 km/h
   t₃= 53.3 min v₃= 42.6 km/h
   d₁= 70.8×28.1/60 = 33.158 km
   d₂= 110×12/60 = 22 km
   d₃= 42.6×53.3/60 = 37.84 km

2. Find d_tot
   d_tot= d₁+d₂+d₃= 33.16+22+37.84= 93.001 km

51.76 km/hr

Explanation:
1. Find average speed: v_tot=d_tot/t_tot
t_tot = (28.1+12+53.3+14.4)/60
divide time (min) by 60 → hr
14.4 min comes from the time stopped/not driving
v_tot = 93.001 km/[(28.1+12+53.3+14.4)/60] = 51.76 km/hr

1. away from PMA at 40 yds/min

Explanation:

1. Find the slope at t=3 using (rise/run)
   (240-160)/(4-2) = 40

10.3 m

Explanation:

1. find t_b for the bird to finish flying the L
   given: v_b = 33.6 km/hr, d = 6 km
   v_b = d_b/t_b → t_b = d_b/v_b = 6/33.6 = 0.1786 hr

2. find x_r at t<sub>b
   - find the distance traveled by the runner during the time it took for the bird to make it to the finish line</sub>
   d_r = v_r*t_r = 5.6 * 0.1786 = 1 km
   6 km - 1km = 5 km left for the runner

3. find closing velocity (v_close)
   v_b + v_r = 33.6 + 5.6 = 39.2 km/hr

4. find closing time (t_close) to travel 5 km
   t_close = d/v_close = 5 km/ 39.2 km/hr = 0.128 s

5. find x_b at t_close
   x_b = v_b*t_close = 33.5 × 0.128s = 4.3008 m

6. find d_total traveled by the bird
   = 6 + 4.3008 m = 10.3 m

36 km

Explanation:
v_b = v_r * 6

• distance traveled by the bird will always be 6x d_r

• d_r will always be 6km

• 6 *6 = 36

9. The diagram describes the acceleration vs. time behavior for a car moving in the x-direction

1. with an increasing velocity

Explanation:

when a is +, v incr

a is -, v decr

10) 2.23×10¹⁴ m/s²
11) 1.07 × 10^-8 m/s

Explanation:
10) what is their a over d = 1.4 cm

1. find a
   given: x = 1.4 cm, v_o = 1 × 10⁵ m/s, v_f = 2.5 × 10⁶
   v² = v_o²+2aΔx → a = (v²-v_o²)/2a = 2.23×10¹⁴ m/s²

11) what is t = ? when the electrons are in the accelerating region

1. find t
   given: x = 1.4 cm, v_o = 1 × 10⁵ m/s, v_f = 2.5 × 10⁶, a = 2.23×10¹⁴ m/s²
   v = v_o + at → t = v-v_o / a = 1.07 × 10^-8 m/s

0.789 s

Explanation:
1. find t for car to reach v_f = 6 m/s
given: v_o = 0, a = 7.6 m/s², v_f = 6 m/s
v = v_o + at → t = v-v_o / a = 0.789 s

25.1 m

Explanation:

given: v_o = 0, v_f = 16.9918 m/s at the end of t = 7.42 s, const acceleration during this period

1. find a during t = 7.42s

v = v_o + at → a = v-v_o / t = 2.24 m/s²
2. find Δx @ t = 4.68202 s
Δx = v_ot + ½at² = 25.1 m
CAN NOT USE Δx = (v_o+v)t/2. WHY? we do not know v_f at t=4.68202 s

EZ

—

Explanation:

A. v_o1 = 14.1 m/s, a₁ = -4 m/s², v_f1= 0 @ red light
v = 0 for t = 36.6 s
v_o2 = 0, a₂ = 3.13, v_f2 = 14.1 m/s

T. const v = 14.1 m/s,

1. find t_a1 for A to stop at light

2. find t_a2 after red light to get to v_f2

3. find t_totalA traveled by A

4. find x_a1 traveled to the light

5. find x_a2 traveled from light to v_f2

6. find x_b using t_totalA

7. find distance between T and A

1. 9.8m/s²

2. Unable to determine

3. less than 9.8 m/s²

4. greater than 9.8 m/s²

1.9.8 m/s²

Explanation:

being thrown downward only charges initial velocity. acceleration will always be 9.8 m/s^2 from gravity.

Rocket

1505.45 m

Explanation:
1. find v_f1 of rocket when it runs out of fuel
given: v_o1 = 0, const a₁ = 29.6 m/s² for t= 5.03 s
v_f1 = v_o + at= 0 + (29.6×5.03) = 148.888 m/s

for step 3

2. find Δx₁ traveled by rocket until it runs out of fuel
   Δx = v_ot + ½at² = ½at² = ½(29.6)(5.03)² = 374.453 m

3. find Δx₂
   given: v₀₂ = v_f1 = 148.888 m/s, a = -9.8 m/s², v_f2 = 0 m/s
   v² = v²_o + 2aΔx_{2 →} Δx₂ = v² -v²_o / 2a = 0-148.888²/(2*-9.8) = 1131 m

4. find x_tot
   374.453 + 1131= 1505.45 m

Ball Roof/window

19.51 m

Explanation:

draw diagram

Start with the window

1. Find v_ow at top of window
   given: v_fw = n/a don’t need, t = 0.124 s, Δx_w = 2.5, a = 9.8 m/s²
   Δx = v_ot + ½at² → v_{o =} Δx-½at² / t = 19.55 m/s

2. Find Δx_r
   given: v_oR = 0 (from rest), a = 9.8 m/s², v_fr = v_ow = 19.55 m/s
   v² = v²_o + 2aΔx → Δx = v² -v²_o/2a = 19.51 m

13.37 s

Explanation:

Look at the 2nd half of the graph

t_A * 2 = total time ball is in the air

@ peak v_oA = 0

v_f = n/a unknown, dont need

1. find t_A
   given: a = 9.8 m/s², Δy = 219 m, v_o = 0
   Δy = v_ot + ½at² → 219 = 0 + ½at² → t = (219/(½a))^1/2 = 6.685 s

2. find t_tot
   t_A x 2 = 6.685 x 2 = 13.37 s

0.4957 s

Explanation:

set 2 equations for b₁ & b₂ together: y_f = y_i + v_ot + ½at²
ground is y = 0
1. ball 1
y_f = y_i + v_ot + ½at² = 17 + 0t + ½(-9.8)t²

2. ball 2

y_f = y_i + v_ot + ½at² = 0 + 35t + ½(-9.8)t²

3. set equal

17 + ½(-9.8)t² = 35t + ½(-9.8)t²
17 = 35t → t = 17/35 = 0.4957 s

10.76 m

Explanation:

1. To find the distance of the sled, take the integral of a(t) to find v(t)
   v(t) = ∫ a(t) dt = (4.7/2)t² + 5.9t + c
   because the rocket starts at rest: v(0) = 0

2. Plug in v(0) = 0 to find the value of C
   v(0)= (4.7/2)0² + 5.90 + c = 0
   c = 0

3. Find x(t) between t=0 & t=1.6s by taking integral of v(t)
   x(t) = ∫^1.6₀ v(t) dt = ((4.7/2)/2)t³ + (5.9/2)t² |^1.6₀
   = [((4.7/2)/2)(1.6)³ + (5.9/2)(1.6)²] - [(4.7/2)/2)0³ + (5.9/2)0²]
   = 10.76 m
   

22) 6.26 m/s

23) -1.08 m/s

24) -8.54 m/s²

Explanations:

y(t) = 6.26t - 4.27 t²

22) To find the ball’s initial speed.

1. Find velocity equation by taking the derivative of y(t)
   v(t) = dy/dt → 6.26 - 8.54t

2. v_i = when t = 0s = v(0)
   v(0) = 6.26 - 8.54×0
   v(0) = 6.26 m/s

23) Find velocity at t = 0.859 s

1. same thing as v(0.859). plug in
   v(0.859) = 6.26 - 8.54×0.859 s = -1.08 m/s

24) Find acceleration at t = 0.859 s

1. Find acceleration equation by taking the derivative of v(t)
   a(t) = dv/dt → - 8.54
   This means acceleration is constant @ -8.54 m/s²