The acceleration of a particle traveling along a straight line is $a = (8 - 2 s) \frac{m}{s^{s}}$ $a = (8-2s) m/s^s$ , where $s$ $s$ is in meters. How do you determine the position of the particle when the velocity is maximum?

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The acceleration of a particle traveling along a straight line is $a = (8 - 2 s) \frac{m}{s^{s}}$ $a = (8-2s) m/s^s$ , where $s$ $s$ is in meters. How do you determine the position of the particle when the velocity is maximum?

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Answer 1 · 2016-05-20T17:50:19

I got at $4 m$ $4m$

Explanation:

I would use Algebra and Kinematics here :
I know that acceleration is the derivative of velocity BUT also that if I want the maximum of a function I need to derive it and set the derivative equal to zero;
here I have the derivative of velocity (acceleration) so I set it equal to zero to get:
$8 - 2 s = 0$ $8-2s=0$
$s = \frac{8}{2} = 4 m$ $s=8/2=4m$

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The acceleration of a particle traveling along a straight line is $a = (8 - 2 s) \frac{m}{s^{s}}$ $a = (8-2s) m/s^s$ , where $s$ $s$ is in meters. How do you determine the position of the particle when the velocity is maximum?

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The acceleration of a particle traveling along a straight line is a = (8-2s) m/s^sa=(8−2s)mssa = (8-2s) m/s^s, where sss is in meters. How do you determine the position of the particle when the velocity is maximum?

1 Answer

Explanation:

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The acceleration of a particle traveling along a straight line is $a = (8 - 2 s) \frac{m}{s^{s}}$ $a = (8-2s) m/s^s$ , where $s$ $s$ is in meters. How do you determine the position of the particle when the velocity is maximum?