A particle travels along a straight line with a velocity of&nbsp;v=(4t−3t2)ms, where t is in seconds. Find the location of the particle when&nbsp;t=4s.&nbsp;s=0&nbsp;when&nbsp;t=0.

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A particle travels along a straight line with a velocity of&amp;nbsp;v=(4t−3t2)ms, where t is in seconds. Find the location of the particle when&amp;nbsp;t=4s.&amp;nbsp;s=0&amp;nbsp;when&amp;nbsp;t=0.

Jonathan Burroughs · Accepted Answer

We denote positive motion is to the right, measured from the origin O for the particle in&amp;nbsp;t=0&amp;nbsp;The position as a function of time can be found by calculating.&amp;nbsp;v=ds&amp;nbsp;dt&amp;nbsp;&amp;nbsp;with&amp;nbsp;t=0,&amp;nbsp;s=0&amp;nbsp;ds=v&amp;nbsp;dt&amp;nbsp;&amp;nbsp;ds=(4t−3t2)&amp;nbsp;dt&amp;nbsp;&amp;nbsp;∫0sds=∫0t(4t−3t2)&amp;nbsp;dt&amp;nbsp;&amp;nbsp;s=(2t2−t3)&amp;nbsp;The position at&amp;nbsp;t=4s&amp;nbsp;After that:&amp;nbsp;s=(2×(4)2−43)=−32m&amp;nbsp;The negative sign indicates that at&amp;nbsp;t=4&amp;nbsp;is the particle is a position 32 m left of the initial position.

Janet Young · Accepted Answer

Step 1  We have  v=4t−3t2  We integrate the velocity equation with respect to t  ∫v dv=∫4t−3t2 dt  s=∫4t dt−∫3t2 dt  s=4[t22]−3[t33]+C  s=2t2−t3+C  We have s=0, when t=0  0=2(0)2−(0)3+C  C=0  s=2t2−t3  Step 2  Plugging t=4 in the position equation, we get  s=2t2−t3  s=2(4)2−43  s=32−64  s=−32units

A particle travels along a straight line with a velocity

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