Explanations & Calculations
- By considering the symmetry of the motion the distance the train travels during the period of acceleration & deceleration should be equal & therefore is 2500km
- Applying s=ut+21at2 for the period of acceleration,
2500kmaaa=0t+0.5×a×(0.5h)2=20000kmh−2=20000×(103)m×(3600s)−2=1.54m−2≈2ms−2
- Applying v=u+at for the period of acceleration
Vmax=0+20000kmh−2×0.5h=104kmh−1≈2.8kms−1
- Applying Newton's second law towards the moving direction during the period of deceleration,
∣F∣=m∣a∣=4.5×105kg×1.54ms−2=693kN
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