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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?

I tried avg accel = delta v over delta t. So, 18-30 / 2.4 = -5.

But the books says, "-20."

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    Change in velocity is $48$ not $12$ – 2017-01-31
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    Note that you were given *speeds* and used them as *velocities*... – 2017-01-31

2 Answers 2

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Average acceleration is Final velocity - Initial velocity / total time. So, $$a=\frac{-30-(18)}{2.4}=-20$$

The need of choosing final and initial velocity as $-30$ and $18$ is because motion is initially in positive direction of $x$.

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    I thought the way it's worded meant 18, then begins going 30 in the opposite direction from that point. – 2017-01-31
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    Is there a way to tell which is which? – 2017-01-31
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    Order doesn't matter so long as you stay consistent and watch your minus signs. – 2017-01-31
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The speed is $v_1=a*t+v_0$, where $v_1=-30m/s$, $v_0=18m/s$ and $t=2.4s$. Then: $$a=\frac{v_1-v_0}{t}=\frac{-30-18}{2.4}\frac{m}{s^2}=-20m/s^2$$