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An object experiences an acceleration of 6.8m/s2.As a result,it accelerates from rest to 24m/s.How much distance did it travel during that acceleration

Answer :

skyluke89

The distance covered by the object is 42.4 m

Explanation:

The motion of the object is a uniformly accelerated motion (at constant acceleration), therefore we can use the following suvat equation:

[tex]v^2 -u^2 = 2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the object in this problem, we have:

u = 0 (it starts from rest)

v = 24 m/s (final velocity)

[tex]a=6.8 m/s^2[/tex]

Solving for s, we find the distance travelled by the object:

[tex]s=\frac{v^2-u^2}{2a}=\frac{24^2-0}{2(6.8)}=42.4 m[/tex]

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