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1. A rock is falling downwards at 12.0 m/s. After 2.0 seconds it is falling at downwards at
31.6 m/s. What was its acceleration?

Respuesta :

hamzaahmeds

Answer:

a = 9.8 m/s²

Explanation:

Here, we will use the first equation of motion to determine the acceleration of the falling rock. The first equation of motion is written as follows:

Vf = Vi + at

where,

Vf = Final Velocity of the falling rock = 31.6 m/s

Vi = Initial Velocity of the falling rock = 12 m/s

a = acceleration = ?

t = time interval = 2 s

Therefore,

31.6 m/s = 12 m/s + a(2 s)

a = (31.6 m/s - 12 m/s)/(2 s)

a = 9.8 m/s²

Cricetus

The acceleration will be "9.8 m/s²".

Given:

  • Final velocity, [tex]V_f = 31.6 \ m/s[/tex]
  • Initial velocity, [tex]V_i = 12 \ m/s[/tex]
  • Time, [tex]t = 2 \ seconds[/tex]

We know the relation,

→ [tex]V_f = V_i +at[/tex]

By substituting the values,

 [tex]31.6 = 12+a(2)[/tex]

     [tex]a = \frac{31.6-12}{2}[/tex]

        [tex]= \frac{19.6}{2}[/tex]

        [tex]= 9.8 \ m/s^2[/tex]

Thus the answer above is correct.

Learn more about acceleration here:

https://brainly.com/question/26104319

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