Answered By Timothy M

We know that dist ance in this case is 50m, so we can substitute that in for d(s). This gives us 50= 0.02(3s²+20s). Now we isolate the s. We move the 0.02 across the equation by dividing both sides by 0.02. 50/0.02 gives us 2500 so we have 2500= 3s²+20s. Quadratic equations require all components to be on one side of the equation and that the equation equals 0. To do this, we subtract 2500 from both sides giving us, 0= 3s²+20s-2500 (or 3s²+20s-2500=0). Now we have a (3), b (20), and c (-2500). We can plug this into the quadratic equation   $s=\left(-b\pm\left(\sqrt{b^2-4ac}\right)\right)\div2a$s=(−b±(√b²−4ac))÷2a  . Doing this provides two answers. 25.73 and -32.39. We know that we can't have negative time in this question, so we find that s =25.73 m/s is the correct answer.

Answered By Majid B

Question:

The safe stopping distance, d, in metres, of a Harold Dobson motorcycle on wet pavement, is given by the function d(s)= 0.02(3s^2 + 20s), where s is the speed of the motorcycle in metres per second. Find the speed at which the safe stopping distance is 50 metres?

Answer:

$d\left(s\right)=0.02\left(3s^2+20s\right)$d(s)=0.02(3s²+20s) 

$d\left(s\right)=50$d(s)=50  

$50=0.02\left(3s^2+20s\right)$‍50=0.02(3s²+20s) 

 $50=\left(0.02\times3s^2\right)+\left(0.02\times20s\right)$‍50=(0.02×3s²)+(0.02×20s)

 $50=0.06s^2+0.4s$50=0.06s²+0.4s 

 $0.06s^2+0.4s-50=0$0.06s²+0.4s−50‍=0 

 $aX^2+bX+c=0$aX²+bX+c=0 

 $X_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$X_1,2=−b±√‍b²−4ac2a   

   $s_{1,2}=\frac{-0.4\pm\sqrt{0.4^2-4\left(0.06\right)\left(-50\right)}}{2\left(0.06\right)}$s_1,2=−0.4±√0.4²−4(‍0.06)(−50)2(0.06) 

 $s_{1,2}=\frac{-0.4\pm\sqrt{12.16}}{0.12}$s_1,2=−0.4±√‍12.160.12 

 $s_{1,2}=\frac{-0.4\pm3.487}{0.12}$s_1,2=‍−0.4±3.4870.12  

   $s_1=\frac{-0.4-3.487}{0.12}=-32.39$s₁=−0.4−3.4870.12 =−32.39  (Unacceptable answer: Negative number)

  $s_2=\frac{-0.4+3.487}{0.12}=25.73$s₂=−0.4+3.4870.12 =25.73 m/s  (Acceptable answer: Positive number)