At a certain time a particle had a speed of 43 m/s in the positive x direction, and 5.8 s later its speed was 95 m/s in the opposite direction. What was the average acceleration of the particle during this 5.8 s interval?

Respuesta :

Answer:

[tex]-23.8\frac{m}{s^{2}} [/tex]

Explanation:

Average acceleration is defined as the change on velocity on an interval of time:

[tex]a_{avg}=\frac{V_f-V_i}{t} [/tex]

with Vf the final velocity, Vi the initial velocity and t the time interval. Note that the initial velocity is positive because is in the positive x direction and final velocity is negative because is in the opposite direction, so:

[tex]a_{avg}=\frac{-95\frac{m}{s}-43\frac{m}{s}}{5.8s} [/tex]

[tex]a_{avg}=-23.8\frac{m}{s^{2}} [/tex]

The negative sign indicates that the acceleration is in the negative x direction.

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