Barbara drove a total of 580.5 580.5 miles in two legs of a trip. In the first part of her trip, she drove at a constant speed for 4.5 4.5 hours. In the second part of her trip, she drove 10 10 miles per hour slower, and drove for 6 6 hours. How fast was Barbara driving in each leg of the trip?

Question

Mathematics

Abram Barr

18 April, 02:25

Barbara drove a total of 580.5 580.5 miles in two legs of a trip. In the first part of her trip, she drove at a constant speed for 4.5 4.5 hours. In the second part of her trip, she drove 10 10 miles per hour slower, and drove for 6 6 hours. How fast was Barbara driving in each leg of the trip?

+1

Answers (1)

Know the Answer?

Ricardo Oconnell · Answer 1

During the first leg, her speed was 61 mph;

During the second leg, her speed was 51 mph

Step-by-step explanation:

In this question, we are told to calculate the speed at which Barbara traveled in both legs of her trip.

Since we do not know the constant speed at which she drove during the first leg of her trip, we can represent this by x. So we can say her speed during the first leg of the trip is x mph.

For the second leg, she drove 10 mph less. Hence her speed here would be (x-10) mph

Let's start by calculating the distance she traveled during the first leg.

Mathematically, distance traveled is = speed * time = x mph * 4.5h = 4.5x miles

For the second part of the trip, the distance traveled is (x - 10) * 6 = (6x - 60) miles

Now the twist to this is that if we add both distances, we get the total distance traveled.

Let's do that;

4.5x + 6x - 60 = 580.5

10.5x = 580.5 + 60

10.5x = 640.5

x = 640.5/10.5 = 61 mph

In the second leg, her speed would be 61 - 10 = 51 mph