Answer:
13 hours
Explanation:
Average speed = Distance traveled / time taken
⇒ Distance = Average speed × Time
d = s × t
For the first trip;
Average speed = 280 mph
d₁ = 280t₁ ------(1)
where;
d₁ is the distance covered to get to the destination
t₁ is the time taken to get to the destination
For the second trip;
Average speed = 240 mph
d₂= 240t₂ ------(2)
where;
d₂ is the distance covered on the way back
t₂ is the time taken on the way back
The trip is the same distance to and fro. Therefore,
d₁ = d₂
Substituting the equation for d₁ and d₂
280t₁ = 240t₂ ------(3)
It took one hour less time to get there than it did to get back, then,
t₁ = t₂ - 1
t₂ = t₁ + 1 ------(4)
Substituting equation (4) into equation (3)
280t₁ = 240(t₁ + 1)
280t₁ = 240t₁ + 240
280t₁ - 240t₁ = 240
40t₁ = 240
t₁ = 240/40
t₁ = 6 hours
From equation (4)
t₂ = t₁ + 1
t₂ = 6 + 1
t₂ = 7 hours
The total time for the trip is t₁ + t₂ = 6 + 7
= 13 hours