Question

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 distance did it travel during that acceleration?

Answer 1

The distance traveled during its acceleration, d = 214.38 m

Step-by-step 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