Goldberg's sleigh currently runs at 203mph, but he needs it to reach 400mph with all the packages he has to deliver.

if goldberg is delivering presents at an altitude of 5,813.6ft, with the same drag and weight of a 2019 challenger srt® hellcat redeye, how much hp would he need to reach 400 mph?

pro tips:
100% driveline efficiency
p (air density) at 5813.6 ft = .0019 slug/ft^3
a (area of challenger front end) = 26.72 ft^2
c(d) (coefficient of drag) = .398
goldberg's sleigh required speed, 400 mph = 587 ft/sec
f(drag) = 1/2 p x v^2 x c(d)a
p = f(drag) x v
convert to horsepower = p ( 1hp/550 ft lbf/sec)

He needed 3747.61 HP to reach 400 mph.

Step-by-step explanation:

Given data :

P (air density)  =  0.0019 slug/ft³

A (Area)  = 26.72 ft²

C(d) = 0.398

crr = 0.015

V = 400 mph = 587 ft/sec = 178.92 m/sec

weight = 2019 lb = 8980.96 N    [1 pound = 4.44822 N]

P =  Force × Velocity = FV

F = F(sum) = F(drag) + F(rolling resistance)

F(drag) =

F(rolling resistance) = weight × crr

  1hp = 745.7 watts

F(drag) =

= 3481.23  

convert    to  

1 slug = 14.5939 kg

1 ft = 0.3048 m

= 14.5939 × 0.3048 = 4.448  or newton

Since F(drag) = 3481.23 × 4.448 N

                      = 15484.51 N

P = F(drag) × V

   = 15484.51 × 178.92 N m/s

   = 2770488.529 watts

F rolling resistance = crr × w

                               = 0.015 × 8980.96 N

                               = 134.714 N

Power = (Fdrag + Frolling resistance) × V

            = (15484.51 + 134.714) × 178.92

            = 15619.224 × 178.92

            = 279459.558 watts

Power in HP

= 3747.61 HP

He needed 3747.61 HP to reach 400 mph.

gregory's graph is incorrect because it shows 2 inches of growth for each week when it should show 0.75 inches of growth each week.

answer: the answer to your question is parallel