Math Question Help?

Question

A stunt pilot is testing a new plane. The equation that models his height over time is h(t) = 15t^2 - 195t + 950, where t is time in seconds and h(t) is his height in meters. Determine when the stunt pilot and the test plane are below 500 meters.

I'm completely lost on how to solve this, please help

Answer 1

Favorite Answer

h (t) = 15t² - 195t + 950

500 >  15t² - 195t + 950

15t² - 195t + 950 - 500 < 0

15t² - 195t + 450 < 0

15 (t² - 13t + 30) < 0

              15 (t - 10)(t - 3) < 0

               when t < 10 and t > 3 .............ANS

Answer 2

h(t)=15t^2-195t+950<500

=>

t^2-13t+30<0

=>

(t-10)(t-3)<0

=>

3<t<10

=>

The plane are below 500 m

when it is more than 3 seconds

& less than 10 seconds after

starting the flight.

Answer 3

We need to find the times when the pilot is at 500 metres.

i.e. when 15t² - 195t + 950 = 500

so, 15t² - 195t + 450 = 0

or, t² - 13t + 30 = 0

Hence, (t - 3)(t - 10) = 0

so, at t = 3 and again at t = 10

As the function is a parabola with a ' ∪ ' shape, the pilot is below 500 metres between these times

i.e. below 500 metres from t = 3 to t = 10 seconds

Hence, 7 seconds in all

:)>

Answer 4

h(t) = 15t² - 195t + 950

We want to find time when h(t) <500

15t² - 195t + 950 < 500

15t² - 195t + 450 < 0

t² - 13t + 30 < 0

(t-10)(t-3) < 0

This gives 2 options

Option-1

(t-10)>0 and (t-3) < 0, this gives...

t>10 and t<3

This is absurd , so ignore

Option-2

(t-10)<0 and (t-3) > 0, this gives...

t<10 and t>3

That means the test plane is below 500m between 3 seconds to 10 seconds

Answer 5

h(t) = 15t^2 - 195t + 950  ((see red graph))

Find  15t^2 - 195t + 950 < 500, so 15t^2 - 195t + 950 - 500 < 0

Find instead where it equals 0

15t^2 - 195t + 450 = 0, solve using Quadratic

At t between 3 & 10, height will be less than 500m

Answer 6

set h(t) = 500 and solve for t values...then check the values of h(t) on either side of those t values