A proton initially moves left to right long the x axis at a speed of 2.00 x 10³ m/s.It moves into an electric field,which points in the negative x direction,and travels a distance of 0.200 m before coming to rest.What acceleration magnitude does the proton experience?

Question

Quizplus · Accepted Answer

The acceleration can be found using the kinematic equation: $ v^2 = u^2 + 2as $. Here, $ v = 0 $, $ u = 2.00 \times 10^3 \, \text{m/s} $, and $ s = 0.200 \, \text{m} $. Solving for $ a $, we get $ a = \frac{-u^2}{2s} = \frac{-(2.00 \times 10^3)^2}{2 \times 0.200} = 1.00 \times 10^7 \, \text{m/s}^2 $.

A Proton Initially Moves Left to Right Long the X