Calculating the time needed until a known velocity for a vehicle with friction as a function of velocity

[Patrick Herp]

Homework Statement

A truck with a mass of 1 ton is travelling at a constant speed of 36 km/h. It is known that the total friction force of all wheels with the asphalt can be expressed by the function $f(v) = 70v + 6v^2$, where v is measured in m/s.
a. Determine the power output of the engine at that moment!
b. The maximum speed of the truck is 108 km/h. Determine the maximum power of the truck's engine!
c. If the truck is initially at rest and the accelerator pedal is pressed as hard as possible, determine the time it takes for the truck to reach half its maximum speed!

Relevant Equations

$$\sum{F} = ma$$

The solutions for (a) and (b) are pretty straightforward, which I got 13 kW and 225 kW each, but when I try to solve for (c), I get stuck with this:
$$
\begin{align}
a &= \frac{F}{m} \nonumber\\
&= \frac{F_\text{max}-f(v)}{m} \nonumber\\
&= \frac{7(30)+6(30)^2 -70v-6v^2}{1.000} \nonumber\\
\frac{dv}{dt} &= \frac{7.500-70v-6v^2}{1.000} \nonumber\\
\frac{dv}{7.500-70v-6v^2} &= \frac{dt}{1000} \implies t = 1.000\int_0^{15} \frac{dv}{7.500-70v-6v^2} \nonumber
\end{align}
$$
Is there any way I could solve for (c) without directly solving that integral?

 

[kuruman]

Why don't you want to solve the integral? Have you not seen the method of partial fractions?

 

[PeroK]

Homework Statement: A truck with a mass of 1 ton is travelling at a constant speed of 36 km/h. It is known that the total friction force of all wheels with the asphalt can be expressed by the function $f(v) = 70v + 6v^2$, where v is measured in m/s.

That looks more like air resistance. The point of having wheels is that they roll and avoid friction. Also, tires are generally made of high-friction rubber, which would be absurd if tire-asphalt friction were constantly acting against a vehicle's motion.

 

[Patrick Herp]

Why don't you want to solve the integral? Have you not seen the method of partial fractions?

For more clarification, I don't really "know" where this question comes from, I just found it randomly on the internet, among other questions that I think belong to high school questions, so I find it really weird that this particular part of the question suddenly jumps in difficulty.
I did get the partial fraction of the integral to this:
$$
\begin{align}
\frac{t}{1.000} &= \int_0^{15} \frac{dv}{7.500 -70v -6v^2} \nonumber \\
&= -\frac{1}{6} \int_0^{15} \frac{dv}{\left( v-30 \right)\left( v+\frac{125}{3} \right)} \nonumber \\
&= -\frac{1}{6}\cdot \frac{3}{215} \int_0^{15} \left( \frac{1}{v-30} - \frac{1}{v+\frac{125}{3}} \right) dv \nonumber \\
\frac{t}{1.000} &= -\frac{1}{430} \left[ \ln{ \frac{v-30}{v+\frac{125}{3}} } \right]_0^{15} \nonumber \\
t &\approx 2.33\text{ s} \nonumber
\end{align}
$$

Basically, I just want to know if there is any way to solve that part without partial fraction or integral of 1/x