Solving Quadratic Word Problems for Jogger and Walker Speeds

Click For Summary
SUMMARY

The discussion focuses on solving a quadratic word problem involving the speeds of a jogger and a walker covering a distance of 5 miles. The jogger travels 1.5 times faster than the walker and finishes 25 minutes earlier. The correct setup involves using the equation r*t=d, where r is the speed of the walker, leading to the equation 5/r = 10/(3r) + 5/12. The conclusion confirms that the jogger's speed is 4 mph and the walker's speed is 6 mph.

PREREQUISITES
  • Understanding of the relationship between speed, time, and distance (r*t=d)
  • Basic algebraic manipulation skills
  • Familiarity with converting minutes to hours for time calculations
  • Knowledge of quadratic equations and their applications
NEXT STEPS
  • Study the derivation of equations from word problems in algebra
  • Learn how to convert time units in mathematical problems
  • Explore more examples of quadratic equations in real-world scenarios
  • Practice solving similar speed and distance problems using algebraic methods
USEFUL FOR

Students learning algebra, educators teaching word problems, and anyone interested in applying mathematical concepts to real-life scenarios involving speed and distance.

kuahji
Messages
390
Reaction score
2
I'm having trouble just figuring out how to set these problems up. Here is one of them.

A jogger and a walker both cover a distance of 5 miles. The runner is traveling 1.5 times faster than the walker and finishes in 25 minutes less time. How fast is each going?

I know it probably should include the r*t=d

So I "tried" to make a table

jogger = rate = 1.5x time = x-25 = 5
walker I had tried x*x+25= 5 but
I then tried to combine the two & got 2.5x^2-12.5x-10=0

I know the answer is 4mph for the jogger & 6 for the walker. Any help on how to get the equation setup or where I'm going wrong would be appreciated.
 
Physics news on Phys.org
"jogger = rate = 1.5x time = x-25 = 5 "
What is x? in "rate= 1.5x" it appears to be the rate (speed) of the walker but in "time= x- 25" it appears to be the time the walker takes to walk the 5 miles. And, if that last "5" is the distance walked, "time= 5" makes no sense. Finally, whatever x is, in x*x+ 25= 5, you can't add 25 minutes to any quantity and get 5 miles as the result.

Let r be the speed of the walker. since "r*t= d", t= d/r so the time required for the walker to go 5 miles is t= 5/r. The speed of the jogger is 1.5r so the time required for the jogger to go 5 miles is t= 5/(1.5r)= 10/(3r). The walker took 25 minutes longer and 25 minutes= 25/60= 5/12 hour so 5/r= 10/(3r)+ 5/12. Multiply the equation by 12r to get rid of the fractions. That does not give a quadratic equation.
 
Thanks, that makes sense. What was throwing me off was the fact that all the other problems in the chapter were quadratic equations. So naturally I wanted to "try" & set the problem up as one... but as you can see I just got confused & "tried" to throw something together that would make it quadratic. This here helped a lot. Thanks again.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
23
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K