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Classical Mechanics Level 2

How long does it take for a sports car with an engine power of 500 hp and a mass of 1.9 tons to accelerate from 0 to 100 km/h?

Details and Assumptions:

Neglect any friction.
The sports car engine has an efficiency of 100%.
A horsepower equals 735.5 watts.

1 second 2 seconds 3 seconds 4 seconds 5 seconds

1 solution

Markus Michelmann
Apr 6, 2018

The sports car accelerates with a constant power of $P = P_0 = 500 \,\text{hp}$ . The work is fully converted into kinetic energy, so that it increases linearly with time: $\begin{aligned} E_\text{kin} &= \frac{1}{2} m v^2 = \int_{0}^t P dt = P_0 t \\ \Rightarrow \quad t &= \frac{m v^2}{2 P_0} = \frac{1900 \cdot (100/3.6)^2}{2 \cdot 500 \cdot 735.5} \,\text{s} \approx 1.99 \,\text{s} \end{aligned}$

If the wheels stick to the ground without limit, the force between the wheels and the track would diverge as $t \rightarrow 0$ . Of course, in real life, the finite friction between the tire and the ground makes the acceleration less dramatic.

One way to think about this is to assume that the force between the wheel and the ground has a maximum cut-off value, for example $F_{max}=mg$ . This corresponds to a coefficient of friction that is larger than one, but that is feasible with the tires of the racing cars. Accordingly, at the beginning the car accelerates with $a=g$ until it reaches the velocity of $v_1=P_0/mg=19.3$ m/s. This takes about $t_1'=v_1/g\approx2$ seconds. Without the upper cut-off in the force the same velocity is reached in about $t_1=\frac{P}{2mg^2}\approx 1$ second. Therefore the limitation of the friction force between the tire and the track adds about 1 second to the total time, increasing it by 50%, from 2 seconds to 3 seconds. Still very fast!

Laszlo Mihaly - 3 years, 2 months ago

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