A car moves along a horizontal circular track of radius R =…

A car moves along a horizontal circular track of radius $R = 43\ \mathrm{m}$ with a constant tangential acceleration $a = 2\ \mathrm{m/s^2}$. Its initial speed is $v_0 = 36\ \mathrm{km/h}$. How long does the first lap take?

Explanation of Method

First lap of a car speeding up on a circular track.

Step 1: Distance along the track

Along the curve, the tangential acceleration changes the speed exactly as in straight-line motion, so we can work with the distance along the track. One lap is

\[ s = 2\pi R = 86\pi\ \mathrm{m} \approx 270.2\ \mathrm{m}, \]

\[ v_0 = 36\ \mathrm{km/h} = 10\ \mathrm{m/s}. \]

Step 2: Equation for the time

\[ s = v_0 t + \tfrac12 a t^2 \;\Longrightarrow\; 270.2 = 10t + t^2, \]

\[ t^2 + 10t - 270.2 = 0. \]

Step 3: Solve

\[ t = \frac{-10 + \sqrt{100 + 4 \cdot 270.2}}{2} = \frac{-10 + 34.36}{2}. \]

(The other root is negative and has no meaning here.)

\[t \approx 12.2\ \mathrm{s}\]

Check

$ 10 \cdot 12.18 + 12.18^2 = 121.8 + 148.4 = 270.2\ \mathrm{m}\ \checkmark $

Speed at the end: $ 10 + 2 \cdot 12.18 \approx 34.4\ \mathrm{m/s} $; average $ \approx 22.2\ \mathrm{m/s} $, and $ 22.2 \cdot 12.18 \approx 270\ \mathrm{m}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $4$ $8$ $12.18$ $10$ $20$ $34.4$ $\text{area} = 2\pi R \approx 270\ \mathrm{m}\ \text{(one lap)}$ $\text{lap done: } 12.2\ \mathrm{s}$ $\text{slope} = a_t = 2\ \mathrm{m/s^2}$

The speed along the track grows steadily from 10 m/s. The lap is finished when the area under the speed–time line equals the length of the track.

Result

\[\boxed{\,t \approx 12.2\ \mathrm{s}\,}\]