Question

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

Answer 1

Answer:

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]

$P=\frac{Force\times Displacement}{Time}$

P =  Force × Velocity = FV

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

F(drag) = $\frac{1}{2}P\times V^2\times C(d)A$

F(rolling resistance) = weight × crr

$HP=\frac{P}{1hp}$   1hp = 745.7 watts

F(drag) = $\frac{1}{2}P\times V^2\times C(d)A$

$=\frac{1}{2}\times 0.0019\times 587^{2} \times 0.398\times 26.72$

= 3481.23  $\frac{slug\times ft}{sec^2}$

convert  $\frac{slug\times ft}{sec^2}$  to  $\frac{kg\times m}{s^2}$

1 slug = 14.5939 kg

1 ft = 0.3048 m

$\frac{slug\times ft}{sec^2}$ = 14.5939 × 0.3048 = 4.448 $\frac{kg\times m}{s^2}$  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

$HP=\frac{P}{hp}$

$=\frac{279459.558}{745.7}$

= 3747.61 HP

He needed 3747.61 HP to reach 400 mph.

   

Answer 2

Answer:

5,813 HP

Step-by-step explanation:

P=f*v

f=1/2*p*v^2*C*A

HP=(f*v)/550