ElectricPavlov

You have the correct answer in METERS.....  Convert to Kilometers and the answer is 'A'

OK    use      vf = vo + at     to find   't'   (note 'a' is negative)

Use this value of 't'     x = xo + vo t + 1/2 a t^2     to find the distance in METERS.....  (xo=0 in this instance)

vf = vo + a t

60 = 50 + (5) t

2=t

I estimate that each pizza costs about $ 10 each    5 pizzas x 10 = $ 50 total       20% would be $10      so $ 5 would be a small tip of 10%

You left off the velocity the rocket has acheived.....I assume it is 50 m/s again

Velocity changed from 0   to 50 m/s     in   t   time

v= vo + at

50 = 0 + at

50/t = a             (Equation 1)

Now we know   x = xo + vo t + 1/2 a t^2   for the thrust portion of the rocket flight   (xo and vo are both ZERO) so we are left with

x= 1/2 a t^2          From Equation 1 :  50/t = a      Substitute this value for a

105 = 1/2 (50/t) t^2

 105 =  25 t

t = 4.2 seconds

That does NOT 'cancel out' ....can yo give mor information?

x^2 - x    =  x (x-1)

=   -8/2187    x   sqrt 2   =    simplest form without converting to decimal

We'll use a couple of different equations regading position and velocity

First:  v = vo + at       vo= 50  a = -9.8 m/s^2    FInd the point where the rocket reaches its apex, v=0

0 = 50 -9.8t      results in  t = 5.1 deconds

How high did it go?

x = xo + vo t + 1/2 a t^2       xo= 105 m    vo = 50 m/s    and   a = -9.8 m/s^2  

   = 105  +  50 (5.1)  - 1/2(9.8) (5.1)^2   = 360-127.45 = 232.55 m

I will assume you mean this:   -8 sqrt(2187)

= -8 sqrt (81*27) = -8 (9) sqrt(27) = -72 sqrt (27) = -72 sqrt(9*3) = -72(3) sqrt (3) = - 216 sqrt3

The distance equals the initial 50 miles at 75 mph + additional miles at 45mph.

50 m / 75mph = 2/3 hr = 40 minutes

The REST of the time was at 45 MPH    2 hr 15 min  - 40 min = 1 hr 35 min   @ 45 MPH

Rate times time = distance        45 MPH x 1 hr 35 min = 45 mph x 1.58333hr = 71.25 miles

PLUS the initial 50 miles = 121.25 miles