Question

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?

Answer 1

Answer:

$-23.8\frac{m}{s^{2}}$

Explanation:

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

$a_{avg}=\frac{V_f-V_i}{t}$

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:

$a_{avg}=\frac{-95\frac{m}{s}-43\frac{m}{s}}{5.8s}$

$a_{avg}=-23.8\frac{m}{s^{2}}$

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