Answer to Question #264049 in Calculus for Lou

Question #264049

A billiard ball is hit and travels in a straight line. If 𝑠 π‘π‘š is the distance of the ball from

its initial position at 𝑑 𝑠𝑒𝑐, then 𝑠 = 100𝑑2 + 100𝑑. If the ball hits the cushion that is 39 π‘π‘š from its

initial position, at what velocity does it hit the cushion? 


1
Expert's answer
2021-11-11T12:26:38-0500

Given:

s(t)=100t2+100ts=39 cm.β‡’39=100t2+100tβ‡’100t2+100tβˆ’39=0β‡’βˆ’100Β±(100)2βˆ’4(βˆ’39)(100)2(100)β‡’βˆ’100Β±10000+15600200=βˆ’100Β±160200=βˆ’100+160200 [t can not be negative]=60200=310s(t) =100 t^{2}+100 t \\ s=39 \mathrm{~cm} . \\ \Rightarrow 39=100 t^{2}+100 t \\ \Rightarrow 100 t^{2}+100 t-39=0 \\ \Rightarrow \frac{-100 \pm \sqrt{(100)^{2}-4(-39)(100)}}{2(100)} \\ \Rightarrow \frac{-100 \pm \sqrt{10000+15600}}{200} \\ =\frac{-100 \pm 160}{200} \\ =\frac{-100+160}{200} \ [\text{t can not be negative}] \\ =\frac{60}{200}=\frac{3}{10}

Velocity (t)=ddt[100t2+100t]=200t+100(t)=\frac{d}{dt}[100t^2+100t]=200t+100\\

At t=310t=\frac{3}{10}

Velocity == 200(310)+100=160 cm/s200(\frac{3}{10})+100=160\ cm/s


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