A particle travels along a straight line with a velocity of v = (20-0.02 s^2) m/s, where s is in...

Question:

A particle travels along a straight line with a velocity of {eq}v = (20-0.02 s^2) \ m/s, {/eq} where {eq}s {/eq} is in meters. Determine the acceleration of the particle at {eq}s = 10 \ m {/eq}.

Velocity:

The mechanical quantity that represents the position cover by the object in unit time in a specific direction is known as velocity. It is an essential part of the kinematic analysis to define the motion of the object.

Answer and Explanation:


Given Data:

  • The velocity of particle is: {eq}v = \left( {20 - 0.02{s^2}} \right)\;{\rm{m/s}} {/eq}


The expression for acceleration of particle is

{eq}{a_p} = v\dfrac{{dv}}{{ds}} {/eq}


Substitute the value and solve the above expression at {eq}s = 10\;{\rm{m}} {/eq}

{eq}\begin{align*} {a_p} &= \left( {20 - 0.02{s^2}} \right)\left( {\dfrac{d}{{ds}}\left( {20 - 0.02{s^2}} \right)} \right)\\ &= \left( {20 - 0.02{s^2}} \right)\left( { - 0.04s} \right)\\ &= - \left( {20 - 0.02{{\left( {10} \right)}^2}} \right)\left( {0.04\left( {10} \right)} \right)\\ &= - \left( {20 - 2} \right)\left( {0.4} \right)\\ &= - 7.2\;{\rm{m/}}{{\rm{s}}^2} \end{align*} {/eq}


Thus the acceleration of particle at {eq}s = 10\;{\rm{m}} {/eq} is {eq}- 7.2\;{\rm{m/}}{{\rm{s}}^2} {/eq} and negative sign shows the deceleration.


Learn more about this topic:

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Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12
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