00:01
So we are told that angela went home from college, traveling at an average speed of 68 .6 miles per hour.
00:16
And so this is from college to home.
00:27
And it took some amount of time, t1.
00:31
And the reverse journey, home to college, she had an average speed of 58 .8 miles per hour, and that took some amount of time, say, t2.
00:50
And in both cases, the total distance traveled is the distance from home to college, which is d doesn't change because it's the same distance, whether you go from college to home or home to college.
01:05
And we're told that the whole trip took nine hours, so we also have a t1 plus t2, which is the length of the total trip, is equal to nine.
01:19
And we want to figure out how much time it took angela to drive from home back to college.
01:25
So we're looking for this t2 here.
01:28
Now we can form two more equations.
01:32
So we know that angela traveled distance deep, from college to home going an average of 68 .6 miles per hour so d must be equal to 68 .6 times the time it took to get there and similarly d must also be equal to i'll put brackets there just so it's clear 58 .8 times t2 the amount of time it took her to go back from home to college so we've got three equations at three unknowns we can solve this.
02:12
So first let's use these two equations for d to get that 68 .6 t1 is equal to 58 .8 t2 and remember we want to solve for t2 so we'd like to isolate it if we can and we know that t1 plus t2 is equal to nine so over here i can say that that t2 is equal to 9 minus t1 and now i can substitute in uh or rather i should do it the other way around because i want to solve for t2 so i can say that t1 is equal to 9 minus t2 and now i can substitute in for t1 here and solve for t2 so 68 .6 times 9 minus t2 is equal to 58 .8 t2.
03:20
So now i'm going to have 68 .6 times 9 minus 68 .6 t2 is equal to 58 .8 t2.
03:40
And now i'll just bring the two terms with t2 to the same side...