Answer:
(a). The magnitude of the acceleration of the proton is 
(b). The initial peed of the protion is 
(c). The time is
.
Explanation:
Given that,
Electric field 
Time = 5.0 sec
Distance 7.00 cm
(a). We need to calculate the acceleration
Using formula of force



Where, E = electric field
m = mass of proton
q = charge of proton
Put the value into the formula


The magnitude of the acceleration of the proton is 
(b). We need to calculate the initial peed
Using equation of motion

Where, s = distance
Put the value into the formula




The initial peed of the protion is 
(c). We need to calculate the time interval over which the proton comes to rest
Using formula

Where, u = initial velocity
a = acceleration
Put the value into the formula


Hence, (a). The magnitude of the acceleration of the proton is 
(b). The initial peed of the protion is 
(c). The time is
.