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Chapter 6

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 .

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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)