At1=0, a proton is projected in the positive x-direction into a region of a uniform electric field of E=-6x1051. The proton travels 7.00 cm as it comes to rest. Determine

Question

a) The acceleration of the proton.
b) Its initial speed,
c) The time interval over which the proton comes to rest
in simple steps

Answer 1

To find the answers to the questions, we can use the kinematic equations of motion.

a) The acceleration of the proton can be found using the equation:

vf^2 = vi^2 + 2aΔx

Since the proton comes to rest, vf = 0. Plugging in the values, we get:

0 = vi^2 + 2a(0.07)

Simplifying the equation, we get:

0 = vi^2 + 0.14a

Since vi = 0 (the proton starts from rest), the equation becomes:

0 = 0 + 0.14a

Simplifying further, we find:

0 = 0.14a

Therefore, the acceleration of the proton is zero.

b) The initial speed of the proton can be found using the equation:

vf = vi + at

Since the acceleration is zero, the equation becomes:

vf = vi

Therefore, the initial speed of the proton is equal to the final speed, which is zero.

c) The time interval over which the proton comes to rest can be found using the equation:

vf = vi + at

Since the acceleration is zero, the equation becomes:

0 = vi + 0t

Simplifying, we find:

0 = vi

Therefore, the time interval over which the proton comes to rest is also zero.

In summary:
a) The acceleration of the proton is zero.
b) The initial speed of the proton is zero.
c) The time interval over which the proton comes to rest is zero.