Question 6.7: A particle P in an inertial reference frame has an initial v...
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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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)