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(D) The velocity $v$ decreases linearly with displacement $s$. The equation of a straight line is $v = ms + c$.At $s=0$,$v=20 \ m/s$,so $c=20$.At $s=3

Vedclass · Accepted Answer

$-20/3 \ m/s^2$

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(D) The velocity $v$ decreases linearly with displacement $s$. The equation of a straight line is $v = ms + c$.At $s=0$,$v=20 \ m/s$,so $c=20$.At $s=30 \ m$,$v=0$,so $0 = m(30) + 20$,which gives $m = -20/30 = -2/3$.Thus,the velocity function is $v = -\frac{2}{3}s + 20$.The acceleration $a$ is given by $a = v \frac{dv}{ds}$.First,find $\frac{dv}{ds} = \frac{d}{ds}(-\frac{2}{3}s + 20) = -\frac{2}{3}$.At $s=15 \ m$,the velocity is $v = -\frac{2}{3}(15) + 20 = -10 + 20 = 10 \ m/s$.Therefore,the acceleration $a = (10)(-\frac{2}{3}) = -\frac{20}{3} \ m/s^2$.

If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $s$ from $20 \ m/s$ to a value approaching zero at $s=30 \ m$,then the acceleration of the particle at $s=15 \ m$ is $........$

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