Answer
a) See the answer below.
b) $159.4\;\rm s,\;478.3\;\rm s$
c) It is not reasonable to neglect kinetic friction.
Work Step by Step
a)
First of all, we need to draw the force diagram of the object, as you see below.
We know that the drag force is given by
$$D=\frac{1}{2}C\rho A v^2$$
This means that the force exerted on the object is not constant and changes with the change of $v$; the drag force is a function of $v$. So we can not use the kinematic formulas.
Thus, the net force exerted on the object horizontally is given by
$$\sum F_x=-D=-\frac{1}{2}C\rho A v_x^2=m\dfrac{dv_x}{dt}$$
whereas $dv_x/dt=a_x$ as a function in time $t$.
$$-\frac{1}{2}C\rho A v_x^2=m\dfrac{dv_x}{dt}$$
Rearranging;
$$-\frac{1}{2}C\rho A dt=m\dfrac{dv_x}{v_x^2 }$$
Integrating;
$$\int_0^t\frac{-1}{2}C\rho A dt=\int_{v_0x}^{v_x}m\dfrac{dv_x}{v_x^2 }=m\int_{v_0}^{v_x}v_x^{-2} dv_x $$
$$\frac{-1}{2}C\rho A t=-m\;\;\left[\dfrac{1}{v_x }\right]_{v_0}^{v_x}=-m\left[\dfrac{1}{v_x }-\dfrac{1}{v_0 }\right]$$
$$\dfrac{ C\rho A t}{2m}=\left[\dfrac{1}{v_x }-\dfrac{1}{v_0 }\right]$$
$$\dfrac{ C\rho A t}{2m}+\dfrac{1}{v_0 }= \dfrac{1}{v_x } $$
$$\dfrac{ C\rho A tv_0+2m}{2mv_0} = \dfrac{1}{v_x } $$
Therefore,
$$v_x =\dfrac{2mv_0} { C\rho A tv_0+2m} $$
We know that $v_0$ is horizontal, so $v_0=v_{0x}$
$$v_x =\dfrac{2mv_{0x}} { C\rho A tv_{0x}+2m} $$
$$\boxed{v_x =\dfrac{ v_{0x}} { 1+C\rho A tv_{0x}/2m} }$$
b) To find how long the car takes to reach 10 m/s, we first need to solve the previous formula to $t$.
$$ v_{0x}=v_x\left[1+\dfrac{v_xC\rho A tv_{0x}}{2m }\right]$$
$$ \dfrac{v_{0x}}{v_x}-1=\dfrac{ C\rho A tv_{0x}}{2m }$$
Thus,
$$t=2m\left[\dfrac{ \dfrac{v_{0x}}{ v_x}-1}{C\rho A v_{0x}}\right] \tag 1$$
Plugging the given at $v_x=10$
$$t=2\cdot 1500\left[\dfrac{ \dfrac{20}{ 10}-1}{0.35\cdot 1.2\cdot (1.6\cdot 1.4) \cdot 20}\right] =\color{red}{\bf159.4}\;\rm s$$
Plugging the given at $v_x=5$
$$t=2\cdot 1500\left[\dfrac{ \dfrac{20}{ 5}-1}{0.35\cdot 1.2\cdot (1.6\cdot 1.4) \cdot 20}\right] =\color{red}{\bf 478.3}\;\rm s$$
c)
If we assumed that the friction between the car and the ice is just the rolling friction. Let's take the rolling coefficient of friction between tires and concrete in table 6.1, $\mu_r=0.02$.
Let's count the friction force now, which is given by
$$f_r=\mu_rF_n=\mu_rmg=0.02\cdot 1500\cdot 9.5=\bf 294\;\rm N$$
Now let's compare it with the drag force at the initial speed of $v_x=20$ m/s.
$$D=\frac{1}{2}C\rho A v_x^2=\frac{1}{2}\cdot 0.35\cdot 1.2\cdot (1.6\cdot 1.4)\cdot 20^2=\bf 188\;\rm N$$
And at $v_x=10$ m/s;
$$D=\frac{1}{2}C\rho A v_x^2=\frac{1}{2}\cdot 0.35\cdot 1.2\cdot (1.6\cdot 1.4)\cdot 20^2=\bf 47\;\rm N$$
In both cases, the friction force is too larger than the drag force.
So it is not reasonable to neglect kinetic friction.
