How Far Should the Runner Sprint to Finish a 10km Race in Under 40 Minutes?

  • Thread starter frog210293
  • Start date
In summary: No, no, that's not what I was getting at. Let me rewrite the model:(time it takes to run x meters @ 3.7 m/s) + (time it takes to run ? meters @ 7.8 m/s) = (total time)Given that t = d/r, the time it takes to run x meters @ 3.7 m/s would be\frac{x}{3.7}The time it takes to run ? meters @ 7.8 m/s would be\frac{\text{?}}{7.8}The total time is 40 min, or 2400 sec, so the equation becomes\
  • #1
frog210293
12
0

Homework Statement



A runner is planning for a 10km race. She can maintain a steady speed of 3.7 m/s for as much time as needed before ending the race with a 7.8 m/s sprint. If she wants to finish in 40min or less, how far from the finish should she begin to sprint?

Homework Equations


S = d / t


The Attempt at a Solution



10 000m - distance
3.7m/s - speed

10 000/3.7
=2702.702s - time it would take if only running 3.7m/s

40*60= 2400s - time needed to complete the race in

2400s - 2702.702s
=302.702s

=302.7s / 7.8m/s
2361.075m - this answer is wrong. can anyone see what i have done wrong, thanks
 
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  • #2
The method is completely wrong. Moreover you have gone with different dimensions. Dividing time by speed does not give you distance.
 
  • #3
Solve it for some time. That is, assume that after t seconds the girl will sprint. Then solve for t.
 
  • #4
Ok then i didn't think it was going to be quite that easy but couldn't think of any other way to do it. So the equation should look something like t= (2702.702s - 2400s)*7.8m/s or is that basically making the same mistake still?
 
  • #5
This time you are dimensionally correct but conceptually wrong. What is the logic behind it? Can't you follow the more traditional method.
 
  • #6
The equation can be modeled this way:

(time it takes to run @ 3.7 m/s) + (time it takes to run @ 7.8 m/s) = (total time)

Use the equation t = d/r, with the appropriate values or expressions for d and r, for the two spots in the left side of the equation. The spot on the right side, total time, you know. Then solve for the distance. (I'm leaving some things out, I know. I hope you can figure the rest out.)
 
  • #7
ok so we so 2703 + 1282.05 = 3985.05s

then d = s / t
d = 0.474 /3985.05s
d= 10000-9962.6
d= 37.4m from the finish? that doesn't sound right sorry i havn't done any sort of equations in a long time
 
  • #8
Yeah this one is correct.
 
  • #9
ok, thanks a lot otherwise i wouldn't of had a clue how to get this one right
 
  • #10
is this the only answer for this question? its coming up as an incorrect answer sorry
 
  • #11
frog210293 said:
ok so we so 2703 + 1282.05 = 3985.05s

then d = s / t
d = 0.474 /3985.05s
d= 10000-9962.6
d= 37.4m from the finish? that doesn't sound right sorry i havn't done any sort of equations in a long time

No, no, that's not what I was getting at. Let me rewrite the model:

(time it takes to run x meters @ 3.7 m/s) + (time it takes to run ? meters @ 7.8 m/s) = (total time)

Given that t = d/r, the time it takes to run x meters @ 3.7 m/s would be
[tex]\frac{x}{3.7}[/tex]

The time it takes to run ? meters @ 7.8 m/s would be
[tex]\frac{\text{?}}{7.8}[/tex]

The total time is 40 min, or 2400 sec, so the equation becomes
[tex]\frac{x}{3.7} + \frac{\text{?}}{7.8} = 2400[/tex]

Now I leave it to you to fill in the "?". If the runner runs x meters (out of 10km) @ 3.7 m/s, how many meters must she run @ 7.8 m/s?
 
  • #12
I have no idea. now we have 2 variables to try and solve. once the ? is known then the rest is simple enough. As we basically have d/s + d/s and we know the speeds, i only need to find the distance she has to run at 7.8m/s as you said. so she runs x/10km at 3.7ms per second she then needs to make up the rest at 7.8m/s which is the remainder of that amount/10km?
 

What is the basic speed/time/distance equation?

The basic speed/time/distance equation is d = st, which represents the relationship between distance (d), speed (s), and time (t).

How is the basic speed/time/distance equation used?

This equation is used to calculate any of the three variables (distance, speed, or time) when the other two are known. It can also be rearranged to solve for different variables, depending on the situation.

What are the units for each variable in the speed/time/distance equation?

The units for distance are typically in meters (m), speed in meters per second (m/s), and time in seconds (s). However, the units can be changed to fit the specific situation.

What if I have multiple speeds and distances in a problem?

In this case, you can use the average speed for the entire distance by dividing the total distance by the total time. Alternatively, you can use the individual speeds and distances to calculate the total time, and then use that time in the basic equation.

Can the basic speed/time/distance equation be used for objects with changing speeds?

Yes, this equation can be used for objects with changing speeds, but it will only give you the average speed. To calculate the speed at specific points, you would need to use calculus and the concept of instantaneous speed.

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