A proton initially moves left to right long the x axis at a speed of 9.00 × 10³ m/s.It moves into an electric field,which points in the negative x direction,and travels a distance of 0.900 m before coming to rest.What acceleration magnitude does the proton experience?

Question

Quizplus · Accepted Answer

The proton's initial velocity is in the positive x direction, but the electric field is in the negative x direction. Therefore, the electric field will act to slow down the proton until it comes to rest.

We can use the equation:

Vf^2 = Vi^2 + 2ad

where Vf is final velocity (0 m/s), Vi is initial velocity (9.00 × 10^13 m/s), a is acceleration magnitude (unknown), and d is distance traveled (0.900 m).

Solving for a, we get:

a = (Vf^2 - Vi^2) / (2d)

a = (0 - (9.00 × 10^13)^2) / (2(0.900))

a = -4.5 × 10^17 m/s^2

Note that the acceleration is negative because it is in the opposite direction of the proton's initial velocity.

Therefore, the best choice is B, 4.5 × 10^17 m/s^2.

A Proton Initially Moves Left to Right Long the X