An object is moving with initial velocity of 20m/s. It must slow down to a stop in .33 seconds and travel 4m. Write a solution that applies a force to the object to satisfy the expected result.?

You may “assume” linear acceleration because you are asked for “a solution”, not all solutions.

F = m * a
d = v * t To stop in the specified distance and time.
4m = v(m/s) * 0.33s ; v = 12.12 m/s

To decelerate:
#a = (v/t) ; a = ((12.12(m/s))/0.33s) = 36.7 m/s^2#

F = m * a

#F = m * 36.7 m/s^2#

Initial velocity #u=20ms^-1#. Final velocity #v=0#, distance traveled #=4m#.

The applicable kinematic equation for a constant acceleration is
#v^2-u^2=2as#
#=>0^2-20^2=2axx4#
#=>a=-400/8=-50ms^-2#

Let us find out for what time this retarding force is applied.
Using the other kinematic equation
#v=u+at#
we get
#0=20+(-50)t#
#t=20/50=0.4s#
We see that the object takes more time than given in the question to come to stop. This implies that it is not a case of constant deceleration.

Let us assume that the object of mass #m# is acted upon by a time dependent force #F(t)#. From Newton's Second law we get
#F(t)=mxxa(t)# .....(1)
We know that acceleration can be written as
#(dv)/dt=a(t)#
#=>dv=a(t)cdot dt#
Using (1) and Integrating both sides we get
#v(t)=int(F(t))/mcdot dt#......(2)

Suppose the force is given by the expression
#F(t)=L+Kt#
Where #L and K# are constants. (2) becomes
#v(t)=int (L+Kt)/mcdot dt#
#=>v(t)= L/m t+K/mt^2/2+C#
where #C# is constant of integration.
As #v(t)=u# initial velocity at #t=0#
Expression becomes

#v(t)=L/mt+K/(2m)t^2+u# .....(3)
We also know that velocity can be written in terms of displacement #s# as
#(ds)/dt=v(t)#
#=>ds=v(t)cdot dt#
Integrating both sides we get
#s=intv(t)cdot dt#
Using (3) we get
#s=int(L/mt+K/(2m)t^2+u)cdot dt#
#=>s=L/mt^2/2+K/(2m)t^3/3+ut+C_1#
#C_1# is constant of integration. Assuming that displacement is zero at #t=0#, we get #C_1=0# and the expression for displacement becomes
#=>s=L/(2m)t^2+K/(6m)t^3+ut# ......(4)

Inserting given values in (3) and (4) we obtain
#0=L/mxx0.33+K/(2m)(0.33)^2+20# .....(5)
#4=L/(2m)(0.33)^2+K/(6m)(0.33)^3+20xx0.33# ......(6)
To solve for #L and K#, multiply (5) with #0.33/2#, it becomes
#0=L/(2m)xx(0.33)^2+K/(4m)(0.33)^3+20xx0.33/2# .....(7)

Subtracting (7) from (6)
#4=K/(6m)(0.33)^3+20xx0.33-(K/(4m)(0.33)^3+20xx0.33/2)#
#=>4=-K/(12m)(0.33)^3-20xx0.33/2#
#K=-233.7m#
Inserting this value of #K# in (5) we get
#0=L/mxx0.33-(233.7m)/(2m)(0.33)^2+20#
#L=-22.0m#

As such applied force #F(t)=(-22.0m-233.7mt)# N