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