An object starts at 11 m/s with an acceleration of 12 m/s? What is its velocity after 4.5 seconds?
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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:
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:
The minimum stopping distance is 141.78 ft.