A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At 2

Question

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? 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?

Explanation / Answer

a(t) = At²
v(t) = a(t) dt = (A/3)t³ + C and since v(0) = 0, C = 0
x(t) = v(t) dt = (A/12)t^4 + C and since x(0) = 0, C = 0

Let's assume the distance to Alpha Centauri is D. Time to reach it is such that
D = (A/12)t^4
t^4 = 12D / A
t = (12D/A)^(1/4)
and half of the time required to get there is t' = t/2 = ½(12/D)^(1/4)

At A.C.,
v(t) = (A/3)(12D/A)^(3/4)
and at the half-time point
v(t') = (A/3)(½)³(12D/A)^(3/4)
so
v(t) / v(t') = 1 / ½³ = 8
it is going 8x faster at A.C. than it is at the half-time point.

2. x(t) = (A/12)(12D/A) = D by my own construction.
x(t') = (A/12)(12D/A)(½)^4
so
x(t') / x(t) = (½)^4 / 1 = 1/16
At half of the time, the ship is 1/16 of the distance

ratio become 8:1

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