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At a certain time, a particle had a speed of 97 m/s in the positive x-direction, and 2.9 s later its speed was 42 m/s in the opposite direction. What was the average acceleration of the particle during this 2.9 s interval?

Respuesta :

Answer:

 a = -47.93 m/s²

Explanation:

given,

speed of particle in positive x-direction, u = 97 m/s

speed of particle in opposite direction, v = -42 m/s

time = 2.9 s

average acceleration, a = ?

now,

[tex]a = \dfrac{-42-97}{2.9}[/tex]

[tex]a = \dfrac{-139}{2.9}[/tex]

   a = -47.93 m/s²

Hence, the average acceleration is equal to  -47.93 m/s²

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