When unbalanced force is acting on the system of mass then as per Newton's II law we can say

here we have

now since the force and mass both given to us so we can find acceleration of object from this
now when 50 N force is applied on a box then this 50 N force is unbalanced force on it and it will produce acceleration to the box
So here we can say that box will accelerate with constant acceleration and hence its velocity will linearly increase with time.
Now the displacement of the object will be given as

so here the displacement will be a quadratic function of time and increases with time as well
so here overall the motion will be of constant acceleration, uniformly increasing velocity and displacement will be a quadratic function of time
Answer:
14m/s2
Explanation:
initial velocity=2m/s
final velocity =?
time=4s
acceleration=3.0m/s
v=u+at
v=(2)+(3*4)
v=14m/s
by substituting the values in the equation we get the value 14m/s.
from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"
here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion
we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity

v = 15.4 m/s