Question #899fb

1 Answer
Jul 16, 2016

Answer:

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)