Algebra help?

[Mercsgrl]

Hi, I recently went back to school and after 6 years I remember nothing! I have an algebra test tomorrow and Im having a hard time with motion problems...was able to get a little bit of help from another forum (which O shall O remain T nameless /images/graemlins/smile.gif) but Im still not getting it....is there a "motion problems for dummies" book out there anywhere? I know that R*T=D... but in something like this....what do I do? A stunt driver was needed at a production site of a hollywood movie. The average speed of the stunt drivers flight to the site was 150 mph, and the average speed of the return trip was 100 mph. Find the distance of the round trip if the total flying time was 5 hours.
Im sure its something simple that Im just not getting so if anyone could help I would appreciate it.

Thanks
/images/graemlins/heart.gifStef /images/graemlins/heart.gif

[shhhnake_eyes]

since they were equal distances im assuming an easy way to do this is say 150 and 100 = 125mph as the rate and then use the formula. Is this wrong?

[ChromePony]

[ QUOTE ]
since they were equal distances im assuming an easy way to do this is say 150 and 100 = 125mph as the rate and then use the formula. Is this wrong?

[/ QUOTE ]

Yeah this doesnt work because they were traveling each speed for the same distances but different amounts of time so you cant take the simple average.

In this case you can see that if the one way distance is 300 miles then one leg takes 2 hours (150mph) and the other 3 (100mph), so the answer would be 600 miles and average speed is 600/5 =120mph not 125.

If you can't just 'see' the ratios you can set up an equation.

d/150 + d/100 = 5

mult each side by 300 to get integer multiples of d

2d + 3d = 1500
d=300....which is of course one leg of the trip.

[New001]

[ QUOTE ]
since they were equal distances im assuming an easy way to do this is say 150 and 100 = 125mph as the rate and then use the formula. Is this wrong?

[/ QUOTE ]
I don't think that's quite correct.

I'm getting 120mph. Can someone check that?

[shhhnake_eyes]

yea, Chrome already pointed out my ignorance to the subject lol

[Matt R.]

Just as an extension to what chromepony said -- it may be easier to see if you do it this way and break it down one step further.

Let d = vt be the distance it takes to travel to the site.
And D = VT is the distance of the return trip.

We know the total time is t + T = 2 hours (sorry for the use of lower case and capital letters, but it's a pain to type substcripts)

What are the t's in this equation? From the original distance equations they are t = d/v and T = D/V.

Just substitute them in

d/v + D/V = 2 hours

And of course the distance for each leg of the trip is the same so d = D.

Then we just arrive at chromepenny's equation after substituting in the known velocities,

d/150 + d/100 = 2 hours

Then just solve for d.

It may be a little easier to see these things if you set the two equations up separately. Plus it is a good way to practice setting up equations for applied mathematics problems -- they can get a lot more complicated and you often can't just see the ratios in your head.

[Mercsgrl]

thanks /images/graemlins/smile.gif

/images/graemlins/heart.gifStef /images/graemlins/heart.gif

[Homer]

Here's my possibly strange way of doing it:

d1 = d2
150*t1 = 100*t2 (t2 = 5-t1...substitute)
150*t1 = 100*(5-t1)
150*t1 = 500 - 100*t1
250*t1 = 500
t1 = 2 hr

d1 = 150*t1 = 150*2 = 300 mi

d2 is the same, so total distance is 600 mi

(just to check...d2 = 100*(5-2) = 100*3 = 300 mi)