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Question

A taxi traveling at a speed of 38.0 mi/h needs a minimum of 48.0 ft to stop. If the same taxi is traveling 63.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration.

Answer 1

1 mile = 1.6 km

so 38 mi/h = 38 x 1.6 x 1000 / 3600 m/s = 16.9 m/s

63 mi/h = (63/38) x 16.9 = 28 m/s

48 ft = 48 x 12 x 2.54 cm = 1463 cm = 14.63 m

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v² = u² + 2as

v = 0, u = 16.9, s = 14.63 giving a = - 9.76 m/s/s

When u = 28, s = -28²/ 2 / - 9.76 = 40.16 m

This is (40.16 / 14.63) x 48 = 131.8 ft