Question

66 For, $\mathit{t}\mathbf{<}\mathbf{0}$, an object of massexperiences no force and moves in the positive xdirection with a constant speed ${\mathrm{v}}_{\mathrm{i}}$. Beginning at t=0, when the object passes position x=0, it experiences a net resistive force proportional to the square of its speed: ${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{-}\mathbf{mk}{\mathbf{v}}^{\mathbf{2}}\hat{\mathbf{i}}$, whereis a constant. The speed of the object after t=0is given by$\mathbf{v}\mathbf{=}{\mathbf{v}}_{\mathbf{i}}\mathbf{/}\left(1+{\mathrm{kv}}_{i}t\right)$. (a) Find the positionof the object as a function of time. (b) Find the object's velocity as a function of position.

Short answer

(a) The position of the object as a function of time is $\mathrm{x}=\frac{1}{\mathrm{k}}\ln \left(1+{\mathrm{v}}_{\mathrm{i}}\mathrm{kt}\right)$

(b) The objects velocity as a function of time is$\mathrm{v}={\mathrm{v}}_{\mathrm{i}}{\mathrm{e}}^{-\mathrm{kx}}$

Step-by-step solution

Defining velocity and displacement

Velocity is defined as the rate of change of the displacement vectors, velocity can be calculated by differentiating the displacement. Displacement can be calculated by the anti-derivative of the velocity.

$v=\frac{dx}{dt}$

Here x is the displacement, v is the velocity.

$x=\int vdt$

Finding the expression for the position as a function of time

(a)

The expression for the displacement is given by

$x=\int vdt$

Substitute $\frac{v_i}{1+v_{i}kt}$for in the above equation

$\begin{array}{l}x={\int }_0^t{v}_i\frac{dt}{1+v_{i}kt}\\ x=\frac{1}{k}{\int }_0^t\frac{v_{i}kdt}{1+v_{i}kt}\\ x=\frac{1}{k}\ln \left(1+k{v}_{i}t\right)\end{array}$

Therefore, the position of the object as a function of time is$x=\frac{1}{k}\ln \left(1+v_{i}kt\right)$

Finding the object’s velocity as a function of displacement

(b)

The object’s velocity as function of time is given as

$v=\frac{v_i}{1+v_{i}kt}$...... (1)

Now, from the position expression, find the expression for t

$\begin{array}{c}x=\frac{1}{k}\ln \left(1+k{v}_{i}t\right)\\ kx=1+k{v}_{i}t\\ t=\frac{e^{kx}-1}{k{v}_i}\end{array}$

Substitute $\frac{{\mathrm{e}}^{\mathrm{kx}}-1}{{\mathrm{kv}}_{\mathrm{i}}}$for in the equation (1).

$\begin{array}{c}v=\frac{v_i}{1+v_{i}k\left(\frac{e^{kx}-1}{k{v}_i}\right)}\\ v=v_i{e}^{kx}\end{array}$

Therefore, the object’s velocity as a function of time is $v=v_i{e}^{-kx}$.