MHB 2.3.17 Find the average acceleration for this interval.

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The average acceleration of the object between t=5 s and t=8 s is calculated using the formula a_av = (v2 - v1) / (t2 - t1), resulting in -2 m/s². The initial velocity is 5 m/s at t=5 s, and the final velocity is -1 m/s at t=8 s. While the calculation confirms the book's answer of -2 m/s², there is insufficient information to create a detailed graph of acceleration over time. The discussion highlights that only average acceleration can be determined from the given data. Therefore, the focus remains on the average value rather than the specifics of the motion's curve.
karush
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2.3.17 At $t=5\, s$ an object is traveling at $5 \, m/s$.
At $t=8\, s$ its velocity is $-1\, m/s$
(a) Find the average acceleration for this interval.
$$a_{av}=\frac{v_2-v_1}{t_2-t_1}=\frac{\Delta v}{\Delta t}$$
So
$$a_{av}=\frac{-1-5}{8-5}=\frac{-6}{3}=-2 \, m/s$$
book answer $-2\, m/s^2$ok I think this is ok
but again was wondering if a tikx graph can be made or is acceleration a curve?
 
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karush said:
2.3.17 At $t=5\, s$ an object is traveling at $5 \, m/s$.
At $t=8\, s$ its velocity is $-1\, m/s$
(a) Find the average acceleration for this interval.
$$a_{av}=\frac{v_2-v_1}{t_2-t_1}=\frac{\Delta v}{\Delta t}$$
So
$$a_{av}=\frac{-1-5}{8-5}=\frac{-6}{3}=-2 \, m/s$$
book answer $-2\, m/s^2$ok I think this is ok
but again was wondering if a tikx graph can be made or is acceleration a curve?
You don't have enough information to draw the actual curve. Just enough to get the average.

-Dan
 
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