Constant acceleration except time?

AI Thread Summary
The discussion revolves around calculating the ratio of accelerations between two runners, A and B, who run the same distance but at different times. Runner B finishes first, and the ratio of their times is given as 4/5. The correct approach involves using the equation for distance under constant acceleration, leading to the conclusion that the ratio of their accelerations is 0.64 or 16/25. The conversation emphasizes the importance of understanding how variables scale in physics equations and applying logical reasoning to solve problems effectively. Ultimately, mastering the relationship between time and acceleration is crucial for solving such physics problems accurately.
Lori

Homework Statement


upload_2017-12-9_18-32-52.png


Homework Equations



d = vit + 1/2at^2

The Attempt at a Solution



Hey, so what i did was different from the answer key above. Instead of (4/5)^2 i did 1/ (4/5^2) so i got 25/16 as the answer. I'm not sure why tb/ta = 4/5 since 4/5 the time is referring to runner b's time? Why is it combined here?[/B]
 

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Runner B is gets to the finish line first. You are asked for the ratio of runner A's acceleration to that of runner B. Which one accelerated faster?
 
Why did you use 1/(4/5)^2?
 
jbriggs444 said:
Runner B is gets to the finish line first. You are asked for the ratio of runner A's acceleration to that of runner B. Which one accelerated faster?
Runner b because both runners ran at constant acceleration
 
Lori said:
Runner b because both runners ran at constant acceleration
nvmd, i was able to get the answer. I had to solve for a for both runners and plug in time, and then put them in ratio
 
The slow guy took ##T## time. The better one did the same distance ##x## in ##0.8T##. You correctly observed that in this instance, ##x=\frac{1}{2}at^2##. We have two equivalent ##x##'s since they ran the same distance. Let's carefully label our accelerations, ##a_1## for the slow guy and ##a_2## for the fast one. Now

$$\frac{1}{2}a_1 T^2=\frac{1}{2} a_2 (0.8T)^2$$

Which easily simplifies to ##a_1=0.64a_2##. Since we want ##\frac{a_1}{a_2}##, we can just rearrange and find that ##\frac{a_1}{a_2}=0.64=16/25##.

When questions ask for the ratio of the same variable in different conditions, it's very useful to think about how that variable scales in its appropriate equation. For example, since time squared runs proportional to acceleration, this problem becomes immediate. I knew the answer was either ##16/25## or ##25/16## but I used logic to sort out which it would be by considering which person's acceleration was greater. So you need to work on thinking about your variables logically. The biggest piece of advice I can give you for physics is that you should be actively applying "physics thinking" to the algebra part. What I mean is that you can't expect to just set up equations using physics and then only use algebra. This works fine with basic problems, but as you get more advanced you will need to be following your algebra with much more than just a mathematical eye. This will motivate more complicated solutions, especially when you're struggling to find useful substitutions.
 
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