Question

Distance of an object question?

Answer 1

if I'm not mistaken \(\bf d(t) = v_ot+\cfrac{1}{2}at^2\\ \quad \\
v_o = \textit{initial velocity} = 12\\
a = \textit{accelaration} = 1.6 = \cfrac{16}{10}\implies\cfrac{8}{5}\\ \quad \\
d(t) = 12t+\cfrac{1}{2}\cdot\cfrac{8}{5}t^2\\ \quad \\
\textit{when }d(t) = 40\\ \quad \\
40 = 12t+\cfrac{1}{2}\cdot\cfrac{8}{5}t^2\implies0 = 12t+\cfrac{1}{2}\cdot\cfrac{8}{5}t^2-40\\ \quad \\
\cfrac{8}{10}t^2+12t-40 = 0\implies
5\left(\cfrac{4}{5}t^2+12t-40\right) = 5(0)\\ \quad \\ 4t^2+60t-200=0\implies t^2+15t-50=0\)

Answer 2

I'd think you'd need to use the quadratic formula for that, doesn't seem to factor out simple enough

Answer 3

Wouldn't you substitute in 40 for the t in the equation because that is the equation for d(t) and they want you to solve for d(40)?

Answer 4

Also, it is decelerating, not accelerating, so would it be -1.6?

Answer 5

ohhh ..... ahemm.. yes

Answer 6

so you'd end up with \(\bf -t^2+15t-50=0\)

Answer 7

d(t) = 40, just means y = 40, so you make the equation, the distance, set to 40

Answer 8

So wouldn't it be:
\[d(40)=12(40)+\frac{ 1 }{ 2 } \times -1.6(40)^2\]

Answer 9

What I don't understand is that this is on a Zeros of Polynomial Functions worksheet so i don't see how this problem has to do with that.

Answer 10

well. "t = seconds"
so d(40) will give us how far it has gone after 40 secs

Answer 11

Yeah so wouldn't you plug in 40 for all the t's in the problem?

Answer 12

hmm yes, if we want to find out how long the object has gone after 40 secs

but what they want to know is, how long it took it to be at 40 meters of distance

is a quadratic equation, so it will yield 2 values, the value as it was accelerating and pass through 40 meters and onwards
and the value when it's decelerating and pass through 40meters down to a full stop

Answer 13

hmmm ... that sounded ... a bit off

Answer 14

we would get 2 values for "t" or seconds, ... the car only passes through the 40meters once.... so I gather one will be the valid one, the other one will be extraneous

Answer 15

Oh I think I see what you mean now, thank you!

Answer 16

Hmm both solutions are extraneous for some reason

Answer 17

\(\bf -t^2+15t-50=0\\ \quad \\
\text{quadratic formula}\\
t= \cfrac{ - b \pm \sqrt { b^2 -4ac}}{2a}\implies t= \cfrac{ - 15 \pm \sqrt { 15^2 -4(-1)(-50)}}{2(-1)}\\ \quad \\
t = \cfrac{-15\pm\sqrt{225-200}}{-2}\implies t = \cfrac{-15\pm\sqrt{25}}{-2} \implies \cfrac{-15\pm 5}{-2}\)

Answer 18

that'd give you 2 values, of \(\large \cfrac{-15\pm 5}{-2}\implies
\begin{cases}
\cfrac{-10}{-2}\implies 5\\ \quad \\\quad \\
\cfrac{-20}{-2}\implies 10
\end{cases} \)