PHYSICS HIGH SCHOOL

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration.

Answers

Answer 1
Answer:

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\n\nv = 0 mi/h\n\nu = 44 mi/h\n\ns = 0.00833 mi\n\n=> 0^2 = 44^2 + 2 * a * 0.00833\n\n=> 1936 = -0.01666a\n\na = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\n\nv = 0 mi/h\n\nu = 79 mi/h\n\na = -116206.48 mi/h\n\n=> 0^2 = 79^2 + (2 * -116206.48 * s)\n\n6241 = 232412.96s\n\ns = (6241)/(232412.96) \n\ns = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

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