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Zanzabum
3 years ago
5

An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista

nce did it travel during that acceleration?
Physics
1 answer:
inessss [21]3 years ago
3 0

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

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The difference in time between the two bangs is 1 s.

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Answer:

I = 1.38 A

Explanation:

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