SOLUTION: A driver sets out on a journey. For the first half of the distance, she travels at 30 miles per hour. For the second half of the distance, she travels at 60 miles per hour. What is

Question 102118: A driver sets out on a journey. For the first half of the distance, she travels at 30 miles per hour. For the second half of the distance, she travels at 60 miles per hour. What is her average speed on this trip?
I know that speed=distance/time
so, 30=distance/t1 and 60=distance/t2
but i'm not sure where to go from there.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
YOU ARE ON THE RIGHT TRACK!!!! SOLVE YOUR EQUATIONS FOR t1 AND t2; ADD THEM TO GET TOTAL TIME AND OUR CALCULATIONS BEGIN TO AGREE.
distance(d) equals rate(r) times time(t) or d=rt;r=d/t and t=d/r
Let S=average speed on this trip
S=(Total distance travelled)/(Total time required)
(Note: Here's another problem that sort of underscores what average speed is all about: "You are going on a 60-mile trip. The first 30 miles, you travel at the rate of 30 mph. How fast must you travel the remaining 30 miles in order to average 60 mph???" ANSWER: Impossible!!! Why??? In order to average 60 mph, you must travel the entire 60 miles in an hour. Unfortunately, you have already used up that hour in the first 30 miles.)
Now back to your problem:
Let d=total distance travelled
Let d1=distance travelled for the first half of the trip
And let d2=distance travelled for the second half of the trip
*****Now we know that d1=d2. So total distance (d)=d1+d2=2d1 (or 2d2)
Let t=total time required
Let t1=time to travel the first half and this equals d1/30
Let t2=time to travel the second half and this equals d1/60
*****So Total time (t)=t1+t2=(d1/30+d1/60)=(2d1/60+d1/60)=(3d1/60)=d1/20
So S=2d1/(d1/20)---Now we multiply numerator and denominator by (20/d1) to get rid of the complex fraction and we get:
S=(2d1(20/d1))/(d1/20*20/d1)=(40d1/d1)/1 or 40 mph (the d1's cancel)
S=40 mph-------------------------------ans

Hope this helps---ptaylor
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