Speed-distance Algebra 1 Problem

In summary, the conversation is about a student struggling with a math problem involving the speeds of a cheetah and a gazelle. The student is unsure of how to represent the problem algebraically and is seeking help from others. The conversation also includes attempts at solving the problem and clarification from other users.
  • #1
Wolfowitz
12
0
Toward the end of Middle School, I always knew it would come time for me to confront my algebraic innumeracy by means of rigorous and meticulous study of textbooks made available to me by friends who have taken Algebra 1 and Algebra 2 courses, to acquaint myself with what I need to know in order to take a set of courses that are math-intensive. I ask that you squelch whatever thoughts you have about me, a Grade 11 student who should be able to solve such problems without worry.

Below is the question that stultifies me, accompanied by my uneasy undertaking to solve it.

Homework Statement


2gtw0sl.png

Homework Equations


s = d/t, I imagine.
3. Attempt at a solution

a. I'm confident with my response to this one:

47x, 'x' being the number of hours the gazelle spent travelling: the totality of this expression upon the value of 'x' being discerned would represent distance travelled.

b. I'm confident with my response to this one, too:

65x, 'x' being the number of hours the cheetah spent travelling.

c. Here's where it gets tricky (for me). What I got was:

65x - 1

My reasoning, simple: it takes x hours for the cheetah to catch the gazelle. I don't exactly know how the -1 comes into play here (and not +1, considering he has to run another mile).

d. 65x = 47x + 1
[Subtract 47x from both sides]
18x = 1
x = 1/18

It would take 1/18 of an hour for the cheetah to catch up to the gazelle.

If we take what I initially suspected the answer to 'c' to be, we get a negative yielding, which is totally implausible:

65x + 1 = 47x
65x = 57x - 1
18x = -1
x = -1/18

Am I even on the right track?
 
Physics news on Phys.org
  • #2
"My reasoning, simple: it takes x hours for the cheetah to catch the gazelle. I don't exactly know how the -1 comes into play here (and not +1, considering he has to run another mile)."

This is wrong. The cheetah travels 65x miles in time x, while the gazelle travels 47x. As the cheetah approaches the gazelle, the gazelle will be going forward as well. This means that the cheetah won't run for 1/65 hours as you would expect. It will run more. Can you think how much it would run?
 
  • #3
If the gazelle runs at 47 mi/hr and the cheetah runs at 65 mi/h, how fast is the distace between them closing?
 
  • #4
Millennial: I fully understand the gazelle will be running too. If I didn't, I would have put for "c." that the cheetah would have to travel only one mile in order to catch up to the gazelle. lol

Here is my second attempt. (HallsofIvy omits that there is a 1 mile disparity, and doesn't address my unease with representing this algebraically.)

The distance between them closes when the difference between the respective distances traveled is 0:

65*x - 47*x + 1 = 0
18*x + 1 = 0
18*x = -1
x = -1/18

In -1/18 of an hour, the cheetah will catch up to the gazelle.

Hardly a plausible answer.

@RV: I wasn't complaining. It was merely a statement of fact.
 
Last edited:
  • #5
HallsofIvy said:
If the gazelle runs at 47 mi/hr and the cheetah runs at 65 mi/h, how fast is the distace between them closing?

That is the right approach. The rate of approach is 65-47 miles per hour. The distance between them is decreasing by 18 miles per hour. How far from each other are they at the start of part c. ? How much time would be required to close the distance to 0? Now, how far was this for the cheetah, using the actual fastest speed for the cheetah?
 
  • #6
Wolfowitz said:
Millennial: I fully understand tha gazelle will be running too. If I didn't, I would have put for "c." that the cheetah would have to travel only one mile in order to catch up to the gazelle. lol

Here is my second attempt. (HallsofIvy omits that there is a 1 mile disparity, and doesn't address my unease with representing this algebraically.)

The distance between them closes when the difference between the respective distances traveled is 0:

65*x - 47*x + 1 = 0
18*x + 1 = 0
18*x = -1
x = -1/18

In -1/18 of an hour, the cheetah will catch up to the gazelle.

Hardly a plausible answer.

HallsofIvy did not omit anything; he simply left it up to you to do some of the work (as per Forum rules). Also: people cannot tell what makes you feel uneasy if you don't tell them!

RGV
 
  • #7
Clue: The TIME is the same but the distances are different.
 

1. What is the formula for calculating speed in a distance algebra 1 problem?

The formula for calculating speed is speed = distance / time. This means that to find the speed, you must divide the distance traveled by the time it took to travel that distance.

2. How do I solve for distance in a speed-distance algebra 1 problem?

To solve for distance, you can use the formula distance = speed x time. This means that to find the distance, you must multiply the speed by the time it took to travel that distance.

3. What is the difference between speed and velocity in a distance algebra 1 problem?

Speed is the rate at which an object is moving, while velocity is the speed and direction of an object's motion. In a distance algebra 1 problem, speed is usually the focus, but velocity can also be calculated by including the direction of motion in the problem.

4. Can I use any unit of measurement for speed and distance in a distance algebra 1 problem?

Yes, as long as you are consistent with your units throughout the problem. For example, you can use miles and hours for a distance algebra 1 problem, but you cannot use miles and minutes without converting the minutes to hours.

5. How can I check my work when solving a speed-distance algebra 1 problem?

You can check your work by plugging your calculated values back into the original formula. For example, if you solved for speed, you can plug the speed, distance, and time values into the formula speed = distance / time to make sure they all match up. You can also use estimation to see if your calculated values are reasonable based on the given information in the problem.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
24
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
6K
  • General Math
2
Replies
44
Views
3K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
558
  • STEM Academic Advising
Replies
12
Views
3K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
Back
Top