Patrick Herp
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- 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?
$$
\begin{align}
a &= \frac{F}{m} \nonumber\\
&= \frac{F_\text{max}-f(v)}{m} \nonumber\\
&= \frac{7(30)+6(30)^2 -70v-6v^2}{1.000} \nonumber\\
\frac{dv}{dt} &= \frac{7.500-70v-6v^2}{1.000} \nonumber\\
\frac{dv}{7.500-70v-6v^2} &= \frac{dt}{1000} \implies t = 1.000\int_0^{15} \frac{dv}{7.500-70v-6v^2} \nonumber
\end{align}
$$
Is there any way I could solve for (c) without directly solving that integral?
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