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## Q. 6.7

A particle P in an inertial reference frame has an initial velocity $\mathbf{v}_0$  at the place  $\mathbf{x}_0$,  and subsequently moves under the influence of a force that is proportional to the time and acts in a fixed direction e. Find the position and velocity of P at time t .

## Verified Solution

The force on P is given by  $\mathbf{F}(P, t)=k t \mathbf{e}$ , where k is a constant and e is a constant unit vector. Use of this relation in (6.1) and integration of the result as shown in (6.20) with the initial value   $\mathbf{v}(P, 0)=\mathbf{v}_0$  yields the velocity  $\mathbf{v}(P, t)=k t^2 / 2 m \mathbf{e}+\mathbf{v}_0$.  With the initial value  $\mathbf{x}(P, 0)=\mathbf{x}_0$,   a second integration described by (6.21) yields the motion  $\mathbf{x}(P, t)=k t^3 / 6 m \mathbf{e} + \mathbf{v}_0 t + \mathbf{x}_0$.  Let the reader show that if P starts at the origin with velocity $\mathbf{v}_0=v_0 \mathbf{j}$  and the force acts in the direction  $e=i$,  the path of P is a cubic parabola  $x=c y^3$. Identify the constant c.

$\mathbf{F}=\dot{\mathbf{p}}=m \mathbf{a}=m \dot{\mathbf{v}}=m\ddot{\mathbf{x}}$                (6.1)

$\mathbf{v}(P, t)=\frac{1}{m} \int \mathbf{F}(P, t) d t + \mathbf{c}_1$                       (6.20)

$\mathbf{x}(P, t)=\int \mathbf{v}(P, t) d t + \mathbf{c}_2$                 (6.21)