An object with cross section A is shot horizontally across frictionless ice. Its initial velocity is $v_{0x}$at $t_0=0s$. Air resistance is not negligible.

a. Show that the velocity at time t is given by the expression

$v_x=\frac{v_{0x}}{1+C\rho A{v}_{0x}t/2m}$

b. A $1.6-m$-wide, $1.4-m$-high, $1500kg$car with a drag coefficient of $0.35$hits a very slick patch of ice while going $20m/s$. If friction is neglected, how long will it take until the car’s speed drops to $10m/s$? To $5m/s$?

c. Assess whether or not it is reasonable to neglect kinetic friction.

cross section of the object $A$

Initial velocity $v_{0x}$at $t=0s$

velocity at time $t=v_x$

Coefficient of drag $C$

density of air $\rho$

mass of the object$m$

From Newton's second law

$F=ma$

The drag force is

$D=\frac{1}{2}C\rho A(v_x{)}^2$

Consider the negative sign for drag force as it acts opposite to the net force and equal the drag force.

$$\begin{gathered} ma=\frac{1}{2}C\rho A(v_x{)}^2 \\ a=\frac{C\rho A}{2m}(v_x{)}^2 \end{gathered}$$

Here $a=\frac{d{v}_x}{dt}$

Therefore, we get role="math" localid="1650212860312" $$\begin{gathered} \frac{d{v}_x}{dt}=\frac{C\rho A}{2m}(v_x{)}^2 \\ \frac{d{v}_x}{(v_x{)}^2}=\frac{C\rho A}{2m}dt \end{gathered}$$

On integrating the above equation, we get

$$\begin{gathered} {\int }_{v_{0x}}^{v_x}\frac{d{v}_x}{(v_x{)}^2}=\frac{C\rho A}{2m}{\int }_0^{t}dt \\ {\left(\overline{)\frac{-1}{\left(v_x\right)}}\right)}_{v_{0x}}^{v_x}=\frac{C\rho A}{2m}{\left(t\right)}_0^t \end{gathered}$$

$v_x=\frac{v_{0x}}{1+C\rho A{v}_{0x}t/2m}$

Conclusion: The expression for velocity at time $t$has been derived.

length of the car $l=1.6m$

height of the car $h=1.4m$

mass of the car $m=1500kg$

drag coefficient $C=0.35$

Initial velocity role="math" localid="1650213612097" $v_{0x}=10m/s$

Final velocity $v_x=5m/s$

density of air$\rho =1.225kg/m^3$

The velocity expression at time $t$is given by

$v_x=\frac{v_{0x}}{1+C\rho A{v}_{0x}t/2m}$

Plugging all the values in the above equation, we get

$$\begin{gathered} 5=\frac{10}{1+{\displaystyle \frac{(0.35)(1.225)(1.6\times 1.4)\left(10\right)t}{2\times 1500}}} \\ 5=\frac{10}{1+{\displaystyle \frac{(0.35)(1.225)(1.6\times 1.4)\left(10\right)t}{2\times 1500}}} \\ 5=\frac{10}{1+0.0032t} \\ 1+0.0032t=2 \\ t=312.5sec \end{gathered}$$

The time taken for the car to drop its speed$312.5secs.$

If we consider an object that moves by inertia with the drag or frictional force,the net force is not zero and the motion will be decelerated (non-uniform), and we will have to introduce new forces to make it uniform, which is absurd. That is why we neglect the mentioned forces.