A ball rolling down a hill was displaced 19.6 m while uniformly accelerating from rest. If the final velocity was 5.00m/s what was the rate of acceleration

Respuesta :

The rate of acceleration is [tex]0.64 m/s^2[/tex]

Explanation:

Since the ball is moving at constant acceleration, we can solve the problem by using 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 displacement

For the ball in this problem,

u = 0

v = 5.00 m/s

s = 19.6 m

Solving for a, we find the acceleration:

[tex]a=\frac{v^2-u^2}{2s}=\frac{(5.00)^2-0}{2(19.6)}=0.64 m/s^2[/tex]

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