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'It's Only a Minus Sign' printed from https://nrich.maths.org/
Particle A
Qualitative behaviour: moves in positive direction with an
increasing velocity. $$\frac{dx}{dt} = x \Rightarrow \int
\frac{1}{x}dx = \int dt \Rightarrow x = Ae^t$$ $x(0) = 1$ so $A =
1$ and the solution is $x(t) = e^t$.
Particle B
Qualitative behaviour: moves in negative
direction with a decreasing velocity, gradually approching origin.
$$\frac{dx}{dt} = x \Rightarrow x = Ae^{-t}$$ $x(0) = 1$ so $A = 1$
and the solution is $x(t) = e^{-t}$.
Particle C
Qualitative behaviour: moves in
positive direction with an increasing velocity. $$\frac{dv}{dt} = x
\Rightarrow \frac{d^2x}{dt^2} - x = 0 \Rightarrow x = Ae^t +
Be^{-t}$$ If particle starts at origin with velocity $v$ then $x(t)
= \frac{v}{2}(e^t - e^{-t})$.
Particle D
Qualitative behaviour: oscillates
around origin. $$\frac{dv}{dt} = -x \Rightarrow \frac{d^2x}{dt^2} +
x = 0 \Rightarrow x = Ae^{it} + Be^{-it} = C\sin{t} + D\cos{t}$$ If
particle starts at origin with velocity $v$ then $x(t) =
v\sin{t}$.
Particles C and D have symmetrical
equations of motion, in the sense that starting at origin with
negative velocity would result in the same motion but in the
opposite direction.
Observe that if we start particle C
at $+1$ with velocity $-1$ then our equation of motion is $x(t) =
e^{-t}$, so the particle eventially stops at the origin. Are there
any other initial conditions that can be imposed on particles C or
D which force the particle to stop eventually?