A train starts with speed 54 km/h and covers 600 m

A train starts with speed $54\ \text{km/h}$ and covers $600\ \text{m}$ in the first $30\ \text{s}$. Assuming uniform acceleration, find its speed and acceleration at the end of the 30th second. The track has radius of curvature $R$.

Step 1: Convert to SI units

\[ v_{0} = 54\ \frac{\text{km}}{\text{h}} = 54\cdot\frac{1000}{3600}\ \frac{\text{m}}{\text{s}} = 15\ \frac{\text{m}}{\text{s}}. \]

Step 2: Tangential acceleration from the distance travelled

For uniform acceleration $s = v_{0}t+\tfrac12 w_{\tau}t^{2}$: \[600 = 15\cdot30+\frac{w_{\tau}\cdot900}{2} = 450+450\,w_{\tau} \;\Longrightarrow\; w_{\tau} = \frac13\ \frac{\text{m}}{\text{s}^{2}}.\]

Step 3: Speed at $t = 30\ \text{s}$

\[ v = v_{0}+w_{\tau}t = 15+\frac{30}{3} = 25\ \frac{\text{m}}{\text{s}} = 90\ \frac{\text{km}}{\text{h}}. \]

Step 4: Normal acceleration

Motion along a curve of radius $R$ requires a centripetal component directed at the centre of curvature: $w_{n} = \frac{v^{2}}{R} = \frac{625}{R}\ \frac{\text{m}}{\text{s}^{2}}$.

Step 5: Total acceleration

The two components are perpendicular — one along the track, one across it — so they combine by Pythagoras: \[w = \sqrt{w_{\tau}^{2}+w_{n}^{2}} = \sqrt{\frac19+\frac{390\,625}{R^{2}}}\ \frac{\text{m}}{\text{s}^{2}}.\]

Step 6: Verification

\[\text{Mean speed check: } \frac{v_{0}+v}{2}\cdot30 = \frac{15+25}{2}\cdot30 = 600\ \text{m} \;\;\checkmark\]

\[R = 1000\ \text{m}:\ w_{n} = 0.625,\ w = \sqrt{0.1111+0.3906} \approx 0.708\ \frac{\text{m}}{\text{s}^{2}}\]

\[\text{Here the normal part dominates: even a gentle curve bends the train harder than the engine drives it} \;\;\checkmark\]