Question

a proton moving along the x axis has an initial velocity of 4.0x 10^6 m/s and a constant acceleration of 6.0x10^12 m/s^2. what is the velocity of the proton after it has traveled a distance of 80 cm?

Answer 1

distance:

$x=at^2/2+v_0t+x_0$

$0.8=6\cdot10^{12}t^2/2+4\cdot 10^6t$

$3\cdot10^{12}t^2+4\cdot 10^6t-0.8=0$

$t=\frac{-4\cdot10^6+\sqrt{16\cdot10^{12}+9.6\cdot10^{12}}}{6\cdot10^{12}}=1.8\cdot10^{-5}$ s

$v(t)=x'(t)=at+v_0$

$v(1.8\cdot10^{-5})=6\cdot10^{12}\cdot 1.8\cdot10^{-5}+4\cdot 10^6=1.12\cdot10^8$ m/s