At a certain time a particle had a speed of 2

At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval?

Solution:

Let us represent the initial direction of motion as the +x direction. We can also write the average acceleration over a time interval t_{1}\leq t\leq t_{2}:

a_{avg}=\frac{\triangle v}{\triangle t}=\frac{v(t_{2})-v(t_{1})}{t_{2}-t_{1}}

Let us substitute v_{1}=+18 m/s at t_{1}=0 and v_{2}=-30 m/s at t_{2}=2.4 s.

Thus,

a_{avg}=\frac{(-30 m/s)-(+18 m/s)}{2.4 s-0}

a_{avg}=-20 m/s^{2}

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 19.