Answer to Question #122782 in Math for Felix Coleman

Question

(a) A particle P moves along the x-axis with constant acceleration a in the positive x-direction. Initially
P is at the origin and is moving with velocity u in the positive x-direction. Show that the velocity v
and displacement x of P at time t are given by
v = u + at, x = ut +
1
2
at2
,
and deduce that
v
2 = u
2 + 2ax.
(b) The trajectory of a charged particle moving in a magnetic field is given by
r = b cos (Ωt)i + b sin (Ωt)j + ctk,
where b, Ω and c are positive constants. Show that the particle moves with constant speed and find the
magnitude of its acceleration.

Answer 1

Given $x(0)=0, v(0)=u, a(t)=a=const.$

$$a(t)={dv\over dt}=>v(t)=\int adt=u+at$$

$$v(t)={dx\over dt}=>x(t)=\int vdt=\int(u+at)dt=$$

$$=0+ut+{at^2\over 2}=ut+{at^2\over 2}$$

$$v=u+at=>t={v-u\over a}$$

Then

$$x=ut+{at^2\over 2}=u({v-u\over a})+{a\over 2}({v-u\over a})^2=$$

$$={2uv-2u^2+v^2-2vu+u^2\over 2a}={v^2-u^2\over 2a}$$

$$v^2=u^2+2ax$$

(b)

$$r(t)=b\cos(\Omega t) i+b\sin(\Omega t)j+ctk$$

$$v(t)=r'(t)=-b\Omega \sin(\Omega t)i+b\Omega \cos(\Omega t)j+ck$$

$$|v(t)|=\sqrt{(-b\Omega \sin(\Omega t))^2+(b\Omega \cos(\Omega t))^2+(c)^2}=$$

$$=\sqrt{b^2\Omega^2+c^2}=const$$

$$a(t)=r''(t)=v'(t)=-b\Omega^2 \cos(\Omega t)i-b\Omega^2 \sin(\Omega t)j+0k$$

$$|a(t)|=\sqrt{(-b\Omega^2 \cos(\Omega t))^2+(b\Omega^2 \sin(\Omega t))^2}=b\Omega^2$$