College algebra distance question

Question

The motion of a racecar is described by the formula d(t)=30t +10t^2, where d is the distance (in meters) that the car has traveled from its original position and t is the time in seconds since the car started moving. How long, in seconds, will it take for the car to travel 1,800 meters?

(do not use units in your answer)

Answer 1

Set up the quadratic equation

d(t) = 30t+10t²

1800 = 30t+10t²

0 = 10t²+30t-1800

Factoring Method

0 = 10t²+30t-1800

0 = (5t-60)(2t+30)

5t-60 = 0

5t = 60

t₁ = 12

2t+30 = 0

2t = -30

t₂ = -15

Quadratic Formula Method

0 = t²+3t-180

t = [-b±√(b²-4ac)]/(2a)

t = [-3±√((-3)²-4(1)(-180))]/(2(1))

t = [-3±√(9+720)]/2

t = [-3±√729]/2

t = (-3±27)/2

t₁ = 12, t₂ = -15

Hence, it will take 12 seconds for the car to reach 1,800 meters. Remember that time is always positive, so be sure to exclude any negative solutions for time!

Answer 2

Hi Lillian Y.

Given the following:

d(t)= 30t + 10t² and that d(t) = 1800 you can set your function up

1800 = 30t + 10t²

I would recommend putting it in order according to the highest exponent associated with t.

1800 = 10t² + 30t

Subtract 1800 from both sides of the equation to form a Quadratic

0 = 10t² + 30t - 1800

Divide the entire equation by 10 to give

0 = t² + 3t - 180

Factor your quadratic

0 = (t + 15)(t - 12)

Time is not negative so the solution is t = 12

You can check this by plugging 12 back into your original equation

10(12²) + 30(12)= 10(144) + 30(12) = 1440 + 360 = 1800

There are other ways to do the problem and I hope this one helps if you have any questions please write back.

You can also graph your function at Desmos.com to confirm your answer.

Answer 3

1800 = 30t +10t^2

180 = 3t +t^2

t^2 +3t -180 =0

(t-12)(t+15) = 0

t =12 seconds