A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d

Problem from Quiz 5

A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d.
What is its kinetic energy
when it passing through a distance d₁ (d₁ < d)?

Given Data: M = 15,000 kg

Vₒ = 0

V = 15 m / s

d = 150 m

d₁ = 75 m

K₁ = ?

Solution:

K(kinetic energy) = Kₒ(kinetic energy initial) + W(work) = W

W = Fx

K = W = Fd = ½MV²

F = ½MV² / d

K₁ = W₁ = Fd₁ = ½MV² d₁ / d

K₁ = ½(15,000 kg)(15 m/s)² (75m) / (150m) = 843750 J

Google Calculator Code:

0.5*(15000 kg)*(15 m/s)^2*(75m) / (150m)

Answer: 840 kJ

Physics Problem - Average Power

7.36. The electric motor of a 2-kg train accelerates the train from rest to 1 m/s in 1 s. Find the average power delivered to the train during the acceleration.

Solution

Given Data	Useful formulas
M=2kg V₀=0 V=1m/s t=1s pₐ=?	K=½MV² K₀=0 pₐ=W/t W= K-K₀	pₐ=W/t=(K-K₀)/t=½MV²/t pₐ = ½MV²/t pₐ= ½ (2kg) (1m/s)² /1s = =1 kg m/s²∙m/s=1N∙m/s=1J/s=1W pₐ=1W

7.40. A 1000-kg elevator starts from rest.
It moves upward for 4 s with constant acceleration until it reaches its cruising speed of 2 m/s.

(a) What is the average power of the elevator motor during this period?

(b) What is the motor power when the elevator moves at its cruising speed?

Given Data	Useful formulas
M=1000kg (a) V₀=0 V=2m/s t=4s pₐ=? (b) V=const=2m/s	K=½MV² K₀=0 U=Mgh pₐ=W/t W= =(K+U)-(K₀+ U₀) h=V₀t+at²/2 a=(V-V₀)/t p = F V =  = MgV	W= (K+U)-(K₀+ U₀)= K+U= =½MV²+Mgh=½MV²+Mgh h=at²/2=(V/t)∙t²/2=Vt/2 W= ½MV²+MgVt/2 pₐ=W/t=½MV²/t+MgV/2 or pₐ=W/t=M(V/t+g)∙½V ½MV²/t=½(1000kg)(2m/s)²/4s=500W MgV/2=1000kg∙10m/s²∙2m/s /2=10000W pₐ=10500W=10.5kW or pₐ=1000kg∙(2 m/s / 4 s +10 m/s²)∙½∙2 m/s= =10.5kW p=MgV=1000kg∙10m/s²∙2m/s=20000W=20kW

Deadline

The announcement for CityTech students:

The deadline for online quizzes 2,3,4,5, and 6 is October 6 (Thursday), 4:00 PM.

After the deadline, these quizzes will be closed for records, and correct answers will be displayed.

A car moving in the x direction has acceleration aₓ, that varies with time as shown in the figure

A car moving in the x direction has acceleration aₓ, that varies with time as shown in the figure. pic.twitter.com/xLnJUtQIyp

— Physics I (@PHY210) September 28, 2016

Problem from Online Quiz 1

The measured length of a cylindrical laser beam is 12.1 meters, its measured diameter is 0.00121 meters, and its measured intensity is 1.21 × 10⁵ W/m² . Which of these measurements is the most precise?

Solution:
Uncertainty of the measurement result of 12.1 m is 0.05 m so the precision can be shown by the relative error of 0.05m / 12.1m = 0.05/12.1

Uncertainty of the measurement result of 0.00121 m is 0.000005 m so the precision can be shown by the relative error of 0.000005m / 0.00121m = 0.05/12.1

Uncertainty of the measurement result of 1.21 × 10⁵ W/m² is 0.05 × 10⁵ W/m² so the precision can be shown by the relative error of (1.21 × 10⁵ W/m² ) / (0.05 × 10⁵ W/m²) = 0.05/12.1

As relative errors are the same then precision levels are the same.

PHY 215 Quiz 1 (eztestonline.com/695230/14274823527240100.tp4) Is Closed for Records. Nothing Recorded

PHY 215 Quiz 1 (eztestonline.com/695230/14274823527240100.tp4) is closed for records. Nothing recorded.

If a (t) is acceleration and a (t) = a₀ + bt, where a₀ and b are constants, and t is the time

If a (t) is acceleration and a (t) = a₀ + bt, where a₀ and b are constants, and t is the time.

v (t) = v₀ + a₀t + bt² / 2

x (t) = x₀ + v₀t + a₀t² / 2 + bt³ / 6

x (t₁) = x₀ + v₀t₁ + a₀t₁² / 2 + bt₁³ / 6

x (t₂) = x₀ + v₀t₂ + a₀t₂² / 2 + bt₂³ / 6

x (t₂) - x (t₁) = v₀ (t₂ - t₁) + a₀ (t₂² - t₁²) / 2 + b (t₂³ - t₁³) / 6 

Problem and Solution. A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At²...

^{A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At². Assume that the rocket began at rest at the Earth (x = 0) at t = 0. Assuming it simply travels in a straight line from Earth to Alpha Centauri (and beyond), what is the ratio of the speed of the rocket when it has covered half the distance to the star to its speed when it has traveled half the time necessary to reach Alpha Centauri?}

^{a(t) = At²}

^{v(0) = 0}

^{x(0) = 0}

^{D = x(T)}

^{D / 2 = x(τ)}

^{v(τ) / v(½T) = ?}

^Solution

^{a(t) = At²}

^{v(t) = ⅓At³}

^{x(t) = At⁴ / 12}

^{D = AT⁴ / 12}

^{T⁴ = 12D / A}

^{t⁴ = 12x / A}

^{τ⁴ = 12(D / 2) / A}

^{τ⁴ = 12D / A · (½) = (½)T⁴}

^{τ = ⁴√(½) · T}

^{x(τ) = Aτ⁴ / 12 = A(⁴√(½) · T)⁴ / 12 = ½ AT⁴ / 12 = ½D}

^{v(½T) = ⅓A(½T)³ = (½)³ · ⅓AT³}

^{v(τ) = v(⁴√(½) · T) = ⅓A(⁴√(½) · T)³ = (⁴√(½))³ · ⅓AT³}

^{v(τ) / v(½T) = {(⁴√(½))³ · ⅓AT³} / {(½)³ · ⅓AT³} = {(⁴√(½))³ } / {(½)³} = (2 / ⁴√2)³ = 8 / ⁴√8 = (⁴√8)³}

Space station gives physics a boost - Bad Astronomy

Space station gives physics a boost - Bad Astronomy: This is one of the coolest videos I’ve seen in a while: during a routine reboost of the International Space Station to a higher orbit, the astronauts on board show that the station tries to leave them behind! What a fantastic example of Newtons’s First law: an object in motion tends to stay in motion …

Problem: The figure shows the position of a car (black circles) at one-second intervals...

Problem

The figure shows the position of a car (black circles) at one-second intervals.
What is the acceleration at the time t = 4 s?
 

Solution 1:

x(3s) = 24m

x(4s) = 33m

x(5s) = 40m

v(3.5s) ≈ {x(4s) - x(3s)} / 1s ≈ {3m - 24m} / 1s ≈ 9m / s

v(4.5s) ≈ {x(5s) - x(4s)} / 1s ≈ {40m - 33m} / 1s ≈ 7m / s

a(4s) ≈ {v(4.5s) - v(3.5s)} / 1s ≈ {7m / s - 9m / s} / 1s ≈ { - 2m / s} / 1s ≈ - 2m / s²

Solution 2:

a(t) ≈ {v(t + ½Δt) - v(t - ½Δt)} / Δt

v(t) ≈ {x(t + ½Δt) - x(t - ½Δt)} / Δt

v(t + ½Δt) ≈ {x(t+Δt) - x(t)} / Δt

v(t - ½Δt) ≈ {x(t) - x(t-Δt )} / Δt

a(t) ≈ ([{x(t+Δt) - x(t)} / Δt] - [{x(t) - x(t-Δt )} / Δt]) / Δt

a(t) ≈ ({x(t+Δt) - x(t)} - {x(t) - x(t-Δt )}) / (Δt)²

a(t) ≈ {x(t+Δt) - 2x(t) + x(t-Δt )} / (Δt)² derived important formula

Δt = 1s

t = 4s

a(4s) ≈ {x(5s) - 2x(4s) + x(3s)} / (1s)²

a(4s) ≈ {40m - 2 ∙ 33m + 24m} / 1s² ≈ {40m - 66m + 24m} / 1s² ≈ - 2m / 1s² ≈ - 2m / s²

Solution 3:

if a = const

x(t+Δt) = x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t-Δt) = x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t+Δt) + x(t-Δt) = [x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²] + [x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²]

x(t+Δt) + x(t-Δt) = 2x(t) + a ∙ (Δt)²

x(t+Δt) + x(t-Δt) - 2x(t) = a ∙ (Δt)²

x(t+Δt) - 2x(t) + x(t-Δt) = a ∙ (Δt)²

a = {x(t+Δt) - 2x(t) + x(t-Δt)} / (Δt)² derived important formula

Δt = 1s

t = 4s
a = {x(5s) - 2x(4s) + x(3s)} / (1s)²

a = {40m - 2 ∙ 33m + 24m} / 1s² = {40m - 66m + 24m} / 1s² = - 2m / 1s² = - 2m / s²

2 Describing Motion: Kinematics in One Dimension