# Chapter 2 - Motion Along a Straight Line - Problems: 19

acceleration = $20 m/s^2$

#### Work Step by Step

Acceleration is given by $\frac{\Delta v}{\Delta t}$ or in other words, the change in velocity over the amount of time that elapsed. In the question, we are given that the time interval is 2.4 seconds so we need to find the change in velocity. The initial speed is 18m/s, so if we designate this to be in the positive x direction then the velocity is + 18m/s. We are told that 2.4s later, the speed was 30m/s in the OPPOSITE direction (the negative x) so therefore the velocity at this point is -30m/s. The $Delta v = 18m/s - (-30m/s) = 48m/s$ Since this occurs over a period of 2.4seconds then we have 48m/s divided by 2.4s, which gives $20 m/s^2$

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