Question

Question #899fb

Answer 1

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

Explanation:

We know from Newton's Second law of motion that
`vecF=massxxveca`

Also that kinematic equation connecting constant acceleration `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)