Answer:
a) on the way there, the speed = 50 m/hr
b) on the way back home, speed = 30 m/hr
Step-by-step explanation:
[tex]speed (s) = \frac{distance(d)}{time(t)} \\distance = speed\ \times\ time[/tex]
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
