An electron is projected, with an initial speed of vi=1.38e+05 m/sec, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value? Express your answer in meters.

ΔKE=-ΔPE

KE2-KE1= -(PE2-PE1)
ΔKE= KE2-KE1= m(v/2)²/2-mv²/2=
= mv²/8-mv²/2= - 3mv²/8,
ΔPE= PE2-PE1= k•q1q2/r-0=ke²/r
- 3mv²/8 = - ke²/r,
r=8ke²/3mv²= ...
k =9•10⁹ N•m²/C²
e =1.6•10⁻¹⁹ C.
m=9.1•10 ⁻³¹kg