vf=vi+2ad
vf=0; vi=given; d=.2 solve for a.
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?
vf=0; vi=given; d=.2 solve for a.
acceleration = (final velocity - initial velocity) / time
Since the proton comes to rest, its final velocity is 0 m/s.
Given:
Initial velocity (u) = 2.00 * 10^3 m/s
Distance traveled (s) = 0.200 m
Final velocity (v) = 0 m/s
We need to find the time taken (t) to cover the given distance. Rearranging the equation for time:
t = s / u
Substituting the given values:
t = 0.200 m / (2.00 * 10^3 m/s)
t = 0.0001 s
Now we can calculate the acceleration magnitude:
acceleration = (final velocity - initial velocity) / time
acceleration = (0 - 2.00 * 10^3 m/s) / 0.0001 s
acceleration = -2.00 * 10^7 m/s^2
Therefore, the magnitude of the acceleration experienced by the proton is 2.00 * 10^7 m/s^2.
Given:
Initial velocity (u) = 2.00 × 10^3 m/s
Distance traveled (s) = 0.200 m
Final velocity (v) = 0 m/s
The equation that relates these variables is:
v^2 = u^2 + 2as
Substituting the given values, we get:
0^2 = (2.00 × 10^3)^2 + 2a(0.200)
Simplifying the equation:
0 = 4 × 10^6 + 0.400a
Rearranging the equation to solve for the acceleration (a):
0.400a = -4 × 10^6
a = (-4 × 10^6) / 0.400
a = -1 × 10^7 m/s^2
Therefore, the acceleration magnitude experienced by the proton is 1 × 10^7 m/s^2.