FactChecker said:
Averaging two speed numbers is wrong. It depends on how much time was spent at each speed. Average speed would be total distance divided by total time. Use that to figure out how much travel time would give 60mph for the total distance. Then figure out how much time was used in the first 30 mph part.
I guess my train of thought was that she could average 30 mph in any given time interval. For example, driving for only 30 seconds, she could average 30 mph, or driving for an hour could average 30 mph. So the average for the trip then, would be "at what speed was represented on the speedometer most frequently in the trip"?
fresh_42 said:
How long would she have needed for her plan? And where is she after this time?
If I understand your question correctly, she needs and hour for her plan. I feel like I'm supposed to say that at 30 miles, she averaged 30 mph, and she has gone that in distance in
1 hour and thus, her hour is already used up at the 30 mile mark. But then I question that in thinking that she could travel any distance and average 30 miles per hour. Then I come back and say that this is true but she
traveled 30 miles with that average speed, so she traveled for one hour and it's spent. But I do want to say that if she traveled at a high rate of speed, in the 60 miles total, that would bring her average speed for the whole trip up to 60 mph. So let me see if I can piece that together:
She needs an average speed of 60 mph for the
entire trip.
At the 30 mile mark, she averaged 30 mph. And since she averaged 30 mph over a distance of 30 miles, we can say that one hour has passed.
Since one hour has passed, it is impossible for her to average 60 mph
for the entire trip, because she already has traveled for one hour at that was at an average of 30 mph.
So this leads me to ask- if she can't average 60 mph for the whole trip, no matter how fast she drives because the hour has passed, can she make her average speed slower by driving slower? It's probably a stupid question, but it seems like that since 1 hour has passed, that 30 mph average speed is "locked in" in a sense.
MrAnchovy said:
I’ve seen a few of your posts and it seems to me that you have quite a lot of knowledge but you have difficulty applying it to solving problems, particularly word problems. I think this may be because you try to solve problems intuitively, but you haven’t yet had enough practice to develop this intuition. What you should do is use your knowledge to identify relevant equations, translate the words into the symbols of the equation and work through the problem methodically to check the answer you have “guessed”.
So here, you have identified the relevant equation, which you can write as s=dt s = \frac{d}{t} . You know 2 of these values, so write down the value of t t . For the first part of the journey you can write s1=d1t1 s_1 = \frac{d_1}{t_1} and again t1 t_1 is easily found. For this problem, you should not have to go any further.
The key to doing this successfully is writing down what the problem is telling you in equations that you know in order to get closer to the answer. If you find that there are too many unknowns, go back to the question and check you have used all the information (caution: in more advanced exams there may be extra information that you don’t need and will lead you down blind alleys - this is why it is important that you practice solving problems so that you become familiar with what is, and what isn’t, relevant).
I think that's really great advice!
After writing things down, and I'm not sure if I've got the right idea here- ##30~mph=\frac{30}{t}##.
This results in t = 1 hour, so her hour has already passed which, if I even did that correctly, leads me to my previous response.
Those were really great tips and I will definitely practice putting those to use.
fresh_42 said:
maybe the OP finds this useful:
Yes I actually have found that to be an extremely useful resource. That helped me out a lot last semester when I found myself making stupid mistakes on my exams and I still reference it before taking and test or quiz.