Zeros and polynomial functions

Question

An automobile moving at 12 meters per second on level ground begins to decelerate at a rate of 1.6 meters per second squared. The formula for the distance an object has traveled is d(t)=v₀t+ 1/2 at^ 2 , where v₀ is the initial velocity and a is the acceleration. For what value(s) of t does d(t) = 40 meter??

Answer 1

We start with an equation d(t) = V₀t + ^1/₂ at² for distance as a function of time, t.

We are given values for V₀ and a, so let's start by plugging them in:

d(t) = (12)t + ^1/₂ (-1.6)t² = 12t - 0.8t²

And then we are given a value for d(t) of 40, so let's plug that in too:

40 = 12t - 0.8t²

0.8t² -12t + 40 = 0 rearrange to get in standard form of quadratic

=> 4t² - 60t + 200 = 0 multiply to get rid of fraction

=> t² - 15t + 50 = 0 divide by common factor

=> (t - 5)(t - 10) = 0 factorize

hence d(t) = 40 meters at time = 5 seconds, and time = 10 seconds

Answer 2

Using the given formula for distance: d(t) = v₀t + 1/2 at²

We can replace with the given values for distance=40, initial velocity=12, and acceleration=-1.6

40 = 12(t) + 1/2 (-1.6) (t²)

40 = 12t - 0.8t²

We may solve the quadratic equation to equal zero

0.8t² -12t + 40 = 0

Divide each term by 0.8

t² - 15t + 50 = 0

Factoring the quadratic equation:

(t - 10)(t-5) = 0

Solving for t:

t = 10, t = 5. The distance of 40 m is reached when time equals 5 seconds and 10 seconds.