You went to Disney World by car. The trip there took 6 hours. On the way home, people were evacuating for a possible hurricane so your speed decreased by 20 m/hr and it took 10 hours to get home. How fast was the car going on the way there? How fast was it going on the way back home?

a) on the way there, the speed = 50 m/hr

b) on the way back home, speed = 30 m/hr

Step-by-step explanation:

on the way to Disney land:

distance = d

speed = s

time = t = 6 hours

d = s × t

d = s × 6

d = 6s - - - - - (1)

On the way back:

speed decreased by 20m/hr

This means that s₂ = (s - 20)m/hr

time = t₂ = 10 hours

∴ d = s₂ × t₂

d = (s - 20) × 10

d = 10s - 200  - - - - (2)

Note that the distance is the same to and from disney land, therefore, equation (1) equals equation (2)

∴ 6s = 10s - 200

collecting like terms:

10s - 6s = 200

4s = 200

∴ s = 200 ÷ 4

s = 50 m/hr

Therefore on the way there, the car was moving at a apeed of 50m/hr

speed on the way back home (s₂) = (s - 20) m/hr

∴ s₂ = 50 - 20

s₂ = 30 m/hr

Therefore on the way back home the speed was 30 m/hr

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)