Constant acceleration except time?

In summary, the conversation discusses solving a problem involving the ratio of acceleration between two runners. The correct answer is found by considering the relationship between time squared and acceleration. It is important to apply "physics thinking" when using algebra to solve more advanced problems.
  • #1
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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  • #2
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?
 
  • #3
Why did you use 1/(4/5)^2?
 
  • #4
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
 
  • #5
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
 
  • #6
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.
 

Related to Constant acceleration except time?

1. What is constant acceleration?

Constant acceleration refers to the rate of change of velocity over time, where the velocity increases or decreases by the same amount every second.

2. How is constant acceleration different from uniform acceleration?

Constant acceleration refers to a situation where the velocity changes by the same amount every second, while uniform acceleration refers to a situation where the acceleration remains constant over time.

3. How is constant acceleration calculated?

Constant acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the formula a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

4. What is the unit of measurement for constant acceleration?

The unit of measurement for constant acceleration is meters per second squared (m/s²).

5. What are some real-life examples of constant acceleration?

Some real-life examples of constant acceleration include objects falling due to gravity, a car accelerating at a constant rate on a straight road, and a rollercoaster moving along a track with a constant downward slope.

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