A car of mass m = 4000 kg moves at v_0 = 120 km/h on a…

A car of mass $m = 4000\ \mathrm{kg}$ moves at $v_0 = 120\ \mathrm{km/h}$ on a horizontal road. The resistance force is $F_{\text{tr}} = 10\ \mathrm{kN}$. How far does the car travel after the engine is switched off?

Explanation of Method

Coasting distance of a car.

Step 1: Work–energy theorem

The resistance does negative work equal to the kinetic energy lost: $F_{\text{tr}}\,d = \tfrac12 m v_0^2$.

\[ v_0 = 33.33\ \mathrm{m/s}. \]

\[ d = \frac{m v_0^2}{2F_{\text{tr}}} = \frac{4000 \cdot 1111.1}{20\,000}. \]

\[d \approx 222\ \mathrm{m}\]

$v\ (\mathrm{m/s})$ $x\ (\mathrm{m})$ $0$ $100$ $200$ $33.3$ $222\ \mathrm{m}$ $\tfrac12 m v_0^2 = F_{\text{tr}}\,d$

Speed against distance while coasting to a stop. Friction takes away the same kinetic energy on every metre, so $v^2$ falls linearly with distance.

Result

\[\boxed{\,d \approx 222\ \mathrm{m}\,}\]

Remark

Check. The original result is correct. The deceleration is 2.5 m/s², so the car stops in about 13.3 s.