Physical Problem with Two Solutions: Calculus-Based and Physics-Based: v(x)=√(p+qx), x(t)=?

t₀=0, x₀=0, v(x)=√(p+qx). x(t)=?

t₀=0
x₀=0
v(x)=√(p+qx)
x(t)=?

Calculus-based solution:

dx/dt=q⁰·⁵(p/q+x)⁰·⁵

(p/q+x)⁻⁰·⁵dx = q⁰·⁵dt

(p/q+x)⁻⁰·⁵d(p/q+x) = q⁰·⁵dt

2(p/q+x)⁰·⁵ = q⁰·⁵t + c

2(p/q+x)⁰·⁵ -2(p/q+x₀)⁰·⁵= q⁰·⁵(t-t₀)

2(p/q+x)⁰·⁵ -2(p/q)⁰·⁵= q⁰·⁵t

(p/q+x)⁰·⁵ -(p/q)⁰·⁵= ½q⁰·⁵t

(p/q+x)⁰·⁵ = ½q⁰·⁵t+(p/q)⁰·⁵

p/q+x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵ + p/q

x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵

x = p⁰·⁵t + ¼qt²

x = v₀ t + ½at²
v₀=p⁰·⁵
a=½q

Physics-based and algebra-based solution
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)

As it is given:
v=√(p+qx)
so p=v₀², v₀=√p=p⁰·⁵
as x₀=0 then 2a·Δx=2ax
2ax=qx, 2a=q, a=½q

as x = v₀ t + ½at²
then x = √p t + ¼qt²

⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ·ΔΘΣΦΧΨΩαβγδεζηθικλμνξπρςστφχψωϑϓϕ√∛∜∞∠∣∥∫∬∭∮∯≤≥⊥℃ℇ℉ℏ℔№⅀⅏⎛⎞⎜⍊⎝½

Physics Problem with Solution - Acceleration Varies with Time

#Physics #PhysicsProblem #PhysicalProblem #GeneralPhysics #PhysicsI #Physics1 #CalculusBasedPhysics #CalculusBasedPhysicsI #CalculusBasedPhysics1

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What is the acceleration?

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:
t=T-1s
x=v₀t+at²/2
40m=v₀∙4s+a∙(4s)²/2
13m=v₀∙1s+a∙(1s)²/2=(v₀+a∙1s/2)∙1s
v₀=13m/s-a∙1s/2
40m=(13m/s-a∙1s/2)∙4s+a∙(4s²)/2 = 52m-2s²∙a+8s²∙a = 52m+6s²∙a
6s²∙a=-12m
a=-2m/s²

⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙
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Physics Problem. Check again algebraic transformations

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 travelled half the time necessary to reach Alpha Centauri?
Solution:
T - total time necessary to reach Alpha Centauri
D - distance to the star
v - speed
v₁=v(D/2)
v₂=v(T/2)
v(t)=∫At²dt=⅓At³
x(t)=∫v(t)dt=⅓At³dt=⅓¼At⁴
D=AT⁴/12
D/2=AT⁴/24
D/2=At⁴/12; At⁴/12=AT⁴/24; t⁴=T⁴/2; t⁴/T⁴=½; t/T=∜½
v₁/v₂ = v(t)/v(T/2) = {⅓At³} / {⅓A(T/2)³}
=t³ / (T/2)³ =2³ (t/T)³ = 2³ ⨯ (∜½)³ = (2∜½)³ = (∜16⨯∜½)³ = (∜8)³=4∜2
This result is not among the proposed answers to choose from. Check again algebraic transformations.

Physics Problem

Physics Problem