Find the Speed of Each Rocket

[George^Jr]

Two test rockets are fired over a 5600-mile range. One rocket travels twice the speed of the other. The faster rocket covers the distance in two hours less time than the slower. Find the speed of each rocket.

I cannot formulate the equation for solving the above word problem, even after 2 pages of paper and more than a few of hours of attempting. Can someone give me a hint please, and/or even better, point me to an appropriate lesson(s)/article(s)!

I can understand that since rocket 1 is as twice as fast as rocket 2, and since rocket 2 arrives at the destination 2 hours later than rocket 1, that, (I think, anyway)

Distance = 5600 miles

Rocket 1 (Faster Rocket)​	Rocket 2 (Slower Rocket)​
Time = 2 hours	Time = 4 hours
Rate must = 2800 mph	Rate must = 1800 mph

As I can guess the above, so to speak, however, what I cannot "see" or formulate is the equation for solving this problem (if I "guessed" it correctly). Is it possible to solve this problem using a quadratic equation?

This problem is in the chapter "Solving Equations - First Degree and Quadratic" of a Elementary Algebra book (30+ years old), and the chapter does not have the answer or the process for a solution to this particular type of problem, for whatever reason. I suspecting this is more of a "system of two equations in two unknowns" type of problem, which I have not yet studied (getting there).

Not working through this book for school or work, just trying to further (or reacquire) my academic education in my spare time (what little I have, even with the kids gone now) for personal reasons!

Any help will be really appreciated!
Thanks!

 

[Steven G]

Two test rockets are fired over a 5600-mile range. One rocket travels twice the speed of the other. The faster rocket covers the distance in two hours less time than the slower. Find the speed of each rocket.

I cannot formulate the equation for solving the above word problem, even after 2 pages of paper and more than a few of hours of attempting. Can someone give me a hint please, and/or even better, point me to an appropriate lesson(s)/article(s)!

I can understand that since rocket 1 is as twice as fast as rocket 2, and since rocket 2 arrives at the destination 2 hours later than rocket 1, that, (I think, anyway)

Distance = 5600 miles

Rocket 1 (Faster Rocket)​	Rocket 2 (Slower Rocket)​
Time = 2 hours	Time = 4 hours
Rate must = 2800 mph	Rate must = 1800 mph

As I can guess the above, so to speak, however, what I cannot "see" or formulate is the equation for solving this problem (if I "guessed" it correctly). Is it possible to solve this problem using a quadratic equation?

This problem is in the chapter "Solving Equations - First Degree and Quadratic" of a Elementary Algebra book (30+ years old), and the chapter does not have the answer or the process for a solution to this particular type of problem, for whatever reason. I suspecting this is more of a "system of two equations in two unknowns" type of problem, which I have not yet studied (getting there).

Not working through this book for school or work, just trying to further (or reacquire) my academic education in my spare time (what little I have, even with the kids gone now) for personal reasons!

Any help will be really appreciated!
Thanks!

I have a few questions for you to think about.
1st is 2800 really twice 1800?
2nd: Why would you think that Rocket 1 took exactly 2 hours to make this trip? Why not, for example, Rocket 1 took 16 hours and Rocket 2 took 18 hours?
3rd: You found the speeds for each Rocket (just look at your table) so what is the problem??
This is an algebra problem. Are there any unknowns? If yes, then define variable(s) for them. Do you know the total distance? Do you know the speed of Rocket 1 or Rocket 2? Do you know the time it took for each Rocket to get to where they ended up? If the answer is no, then define a variable for these unknowns.

 

[lev888]

What's the formula relating distance, speed and time?
Write it down for both rockets. If a value is not known a variable.
If you are given a relationship between corresponding values you can use the same variable, e.g. x for one rocket, 10x+6 for the other. This is how you reduce the number of variables until, in most cases, you get the same number of variables and equations.
To solve the system of 2 equations and 2 variables use one equation to get one variable on the left hand side and substitute into the second equation.

 

[George^Jr]

I have a few questions for you to think about.
1st is 2800 really twice 1800?

No! That should have been 1400, oops, corrected table:
Distance = 5600 miles

Rocket 1 (Faster Rocket)	Rocket 2 (Slower Rocket)
Time = 2 hours	Time = 4 hours
Rate must = 2800 mph	Rate must = 1400 mph

2nd: Why would you think that Rocket 1 took exactly 2 hours to make this trip? Why not, for example, Rocket 1 took 16 hours and Rocket 2 took 18 hours?

If Rocket 1 (fastest) travels twice (2 times) the speed of Rocket 2 (slowest), then Rocket 1 would have to take half of time that Rocket 2 takes to reach the destination. This half is 2 hours as given in the problem.
So, (in my thinking) Rocket 2 takes twice as long to reach the destination, which is 2 hours longer than Rocket 1. Hence, a 4 hour trip for Rocket 2.

You found the speeds for each Rocket (just look at your table) so what is the problem??

I do not know how to put this into a algebraic equation.
While it works, [MATH]2t(2)=5600[/MATH], this cannot be the proper equation for this type of a problem, I would think.

What's the formula relating distance, speed and time?
Write it down for both rockets. If a value is not known a variable.
If you are given a relationship between corresponding values you can use the same variable, e.g. x for one rocket, 10x+6 for the other. This is how you reduce the number of variables until, in most cases, you get the same number of variables and equations.
To solve the system of 2 equations and 2 variables use one equation to get one variable on the left hand side and substitute into the second equation.

Distance=5600
d=rt

Rocket 1 (Faster Rocket)	Rocket 2 (Slower Rocket)
Time = 5600 / 2r	Time = 5600 / r
Rate = 5600 / (t - 2)	Rate = 5600 / t

From this table, I'd assume,

[MATH]2r(t - 2) = rt [/MATH][MATH]2rt - 4r = rt[/MATH][MATH]2rt - rt = 4r[/MATH][MATH]rt / r =4r / r[/MATH][MATH]t = 4[/MATH]Aha Gee, That was the equation!!

I've spent over 8 hours trying to get to this, gee, and how simple, duh..
Thank you, thank you, thank you Jomo & lev888!!

 

[Steven G]

No! That should have been 1400, oops, corrected table:
Distance = 5600 miles

Rocket 1 (Faster Rocket)	Rocket 2 (Slower Rocket)
Time = 2 hours	Time = 4 hours
Rate must = 2800 mph	Rate must = 1400 mph

If Rocket 1 (fastest) travels twice (2 times) the speed of Rocket 2 (slowest), then Rocket 1 would have to take half of time that Rocket 2 takes to reach the destination. This half is 2 hours as given in the problem.
So, (in my thinking) Rocket 2 takes twice as long to reach the destination, which is 2 hours longer than Rocket 1. Hence, a 4 hour trip for Rocket 2.

Rocket 2 takes twice as long to reach the destination, which is 2 hours longer than Rocket 1---This is Not true at all! Consider this. Suppose it takes you 15 minutes to drive to work. Then you move. So you are saying that it can't take you twice as long to get to work? What if it now takes you 30 minutes to get to work? Since 15 minutes time 2 IS 30 minutes, it does take you twice as long to get to work AND it does NOT take you 2 hours longer to get to work--it only takes you 15 minutes longer!
No suppose it takes me 1 hour to drive to work. Then I get a new work that takes two more hours to drive to work. Therefore it takes me three hours to get to work. Is 3 hours twice 1 hour? From what you said it is (because 3 hours is 2 more hours that 1 hour). Of course it is Not. 3 hours is triple 1 hour.
In the end, if it takes Rocket 2 takes twice as long as Rocket 1 to reach the destination, you can NOT say that it takes 2 hours longer than Rocket 1.
Do you see that?

Onto your reply to my 3rd statement. If you have the answer, then why would you want to use an algebraic equation to get the answer. Just be glad that you have the answer.
I once ask my students (on a dare from another instructor) to give my students the following problem: You have 60 coins consisting of nickels and dimes. You have twice as many dimes as nickels and the total value of these 60 coins is $5.00. How many quarters do you have?
The answer, once you realize it, is quite obvious. You have 0 quarters as the problem stated that you only have nickels and dimes!