To determine the number of seconds, [tex]\( t \)[/tex], it takes for the penny to travel 110 meters, we need to solve the equation for [tex]\( t \)[/tex] when [tex]\( d = 110 \)[/tex] meters. The given equation is:
[tex]\[ d = 3t + 5t^2 \][/tex]
Given [tex]\( d = 110 \)[/tex]:
[tex]\[ 110 = 3t + 5t^2 \][/tex]
Let's rewrite this equation in the standard quadratic form:
[tex]\[ 5t^2 + 3t - 110 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -110 \)[/tex]. To solve for [tex]\( t \)[/tex], we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-3 \pm \sqrt{3^2 - 4(5)(-110)}}{2(5)} \][/tex]
[tex]\[ t = \frac{-3 \pm \sqrt{9 + 2200}}{10} \][/tex]
[tex]\[ t = \frac{-3 \pm \sqrt{2209}}{10} \][/tex]
[tex]\[ t = \frac{-3 \pm 47}{10} \][/tex]
This gives us two solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-3 + 47}{10} = \frac{44}{10} = 4.4 \, \text{seconds} \][/tex]
[tex]\[ t_2 = \frac{-3 - 47}{10} = \frac{-50}{10} = -5 \, \text{seconds} \][/tex]
Since time cannot be negative, we discard the negative solution. Therefore, the time it takes for the penny to travel 110 meters is:
[tex]\[ t = 4.4 \, \text{seconds} \][/tex]
Answer: 4.4 seconds.
