The answer is obvious acceleration lol sorry if it sounds mean
The magnitude of the electric field due to the particle at the given distance is 
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<h3>Electric field strength</h3>
The electric field strength of a charged particle is the force per unit charge in the given field.
The electric field strength of a charge is given as;
where;
- k is Coulomb's constant
- q is the charge
- r is the distance
Thus, the magnitude of the electric field due to the particle at the given distance is .
Learn more about electric field here: brainly.com/question/14372859
Answer: he did travel 15 meters.
Explanation:
We have the data:
Acceleration = a = 1.2 m/s^2
Time lapes = 3 seconds
Initial speed = 3.2 m/s.
Then we start writing the acceleration:
a(t) = 1.2 m/s^2
now for the velocity, we integrate over time:
v(t) = (1.2 m/s^2)*t + v0
with v0 = 3.2 m/s
v(t) = (1.2 m/s^2)*t + 3.2 m/s
For the position, we integrate again.
p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0
Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m
Then the displacement at t = 3s will be equal to p(3s).
p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m
