We have discussed case of constant force and variable force separately.

We know from of motion that ##vecF=massxxveca##

Also that kinematic equation connecting constant ##a## and distance moved ##d## is, taking the scalar parts ##d=ut+1/2at^2## ##=>d=ut+1/2F/mt^2##

As we see that there is linear relation between the Force and distance. Since force is constant, the graph between force and distance traveled is a straight line parallel to the axis depicting Distance.

Make a table showing values of distance traveled in ##t=1,2,3,4,5,6s ” etc.” ## in one column. Second column will contain value of force for all values of ##s##, which is a constant. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

Suppose Force is a function of time ##t## and can be expressed as ##F(t)=m(A t^2+Bt+C)## …….(1) where ##A,B and C## are constants ##=>a(t)=(At^2+Bt+C)## To find velocity ##v(t)## we need to integrate above expression w.r.t time. We get ##v(t)=At^3/3+B t^2/2+Ct+v_@## where ##v_@## is constant of integration and equal to initial velocity.

To find distance traveled ##d(t)## we need to integrate above expression w.r.t time. We get ##v(t)=At^4/12+B t^3/6+Ct^2/2+v_@t+d_@## …..(2) where ##d_@## is constant of integration and equal to initial displacement.

Make a table showing values of distance traveled in ##t=1,2,3,4,5,6s ” etc.” ## as given by equation (2) in one column. Second column will contain corresponding values of force as calculated with equation (1)