Answer to Question #264049 in Calculus for Lou

Question

A billiard ball is hit and travels in a straight line. If 𝑠 𝑐𝑚 is the distance of the ball from

its initial position at 𝑡 𝑠𝑒𝑐, then 𝑠 = 100𝑡2 + 100𝑡. If the ball hits the cushion that is 39 𝑐𝑚 from its

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

Answer 1

Given:

$s(t) =100 t^{2}+100 t \\ s=39 \mathrm{~cm} . \\ \Rightarrow 39=100 t^{2}+100 t \\ \Rightarrow 100 t^{2}+100 t-39=0 \\ \Rightarrow \frac{-100 \pm \sqrt{(100)^{2}-4(-39)(100)}}{2(100)} \\ \Rightarrow \frac{-100 \pm \sqrt{10000+15600}}{200} \\ =\frac{-100 \pm 160}{200} \\ =\frac{-100+160}{200} \ [\text{t can not be negative}] \\ =\frac{60}{200}=\frac{3}{10}$

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

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

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