Question

A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has an initial speed of 6.63×107 m/s and travels 3.63 cm, what then is its speed?

Answer 1

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$