Question

A flying object's height in the sky can be represented by the equation d= 30+vt+id="TexFormula1" title="4t^2\\" alt="4t^2\\" align="absmiddle" class="latex-formula"> where d is the height in feet, v is the velocity in feet per second, and t is the time in seconds. how long has the flying object been traveling if it has a velocity of 8 feet per second and has risen to a height of 126 ft?
a). 4 seconds
b). 6 seconds
c). 8 seconds
d). 16 seconds

Answer 1

Answer:

t=4 seconds

Step-by-step explanation:

The equation given to us is

d=30 + vt + 4t²

d=126 feet

v= 8 ft per second

to find t =?

Now putting the values in the above equation gives

126 = 30 + 8t + 4t²

126 - 30 = 8t + 4t²

96 = 8t + 4t²

Dividing the whole equation by 4

$\frac{96}{4}=\frac{8t}{4}+\frac{4t^{2} }{4}$

it will become

24 = 2t + t²

taking 24 to other sides of equality

0 = 2t + t² -24

t²+2t-24=0

Now we will do mid term breaking of the eqaution

t²+6t-4t-24=0

taking common from the equation

t(t+6)-4(t+6)=0

(t+6)(t-4)=0

so it becomes

t+6 = 0                  or   t-4 =0

t =-6                       or   t = 4 seconds

as time can not be negative

so t=4 seconds