# Question #899fb

Jul 16, 2016

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

#### Explanation:

We know from Newton's Second law of motion that
$\vec{F} = m a s s \times \vec{a}$

Also that kinematic equation connecting constant acceleration $a$ and distance moved $d$ is, taking the scalar parts
$d = u t + \frac{1}{2} a {t}^{2}$
$\implies d = u t + \frac{1}{2} \frac{F}{m} {t}^{2}$

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 $t = 1 , 2 , 3 , 4 , 5 , 6 s \text{ etc.}$ 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
$F \left(t\right) = m \left(A {t}^{2} + B t + C\right)$ .......(1)
where $A , B \mathmr{and} C$ are constants
$\implies a \left(t\right) = \left(A {t}^{2} + B t + C\right)$
To find velocity $v \left(t\right)$ we need to integrate above expression w.r.t time. We get
$v \left(t\right) = A {t}^{3} / 3 + B {t}^{2} / 2 + C t + {v}_{\circ}$
where ${v}_{\circ}$ is constant of integration and equal to initial velocity.

To find distance traveled $d \left(t\right)$ we need to integrate above expression w.r.t time. We get
$v \left(t\right) = A {t}^{4} / 12 + B {t}^{3} / 6 + C {t}^{2} / 2 + {v}_{\circ} t + {d}_{\circ}$ .....(2)
where ${d}_{\circ}$ is constant of integration and equal to initial displacement.

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