Anyone good with math and physics?

[Trav_X]

Found a problem I don't understand. Here is the problem, the characters were originally Hillary and Bill Clinton, but I changed it to make it less political.

 

Runner A beats Runner B by 10 meters in a 100 meter race. Runner B wants to race the 100m again, so Runner A says he will make it fair by starting 10m behind the starting line while Runner B starts at the starting line again. Who will win the second race? Is Runner A giving Runner B a fair chance? How much will the winning runner win by?

 

The question is easy except for one thing: our Physics teacher said that their speed is not linear, but they are actually accelerating constantly for the entire length of the race 

 

I can't figure out how to do this with a quadratic/parabolic graph, which wouldn't change the result for the first race but changes it drastically for the second. Anyone here good at math/physics/problem-solving? Now is the time to impress  

[corrado33]

It's not too bad.

 

In the first race you know that Runner A runs 100m in some time (Tf). Runner B runs 90 meters in the same amount of time (Tf)

 

Using that, you can use the fact that the integral of the V vs. T graph equals the distance traveled. Therefore, if we can model the velocity like so

 

Va = Aa * T (Aa is the acceleration of runner A) and subsequently Vb = Ab * T for runner B.

 

Now, if we integrate (area under the curve) one of those to get the distance traveled we get Distance traveled = Aa ( T^2/2) So, we set the distance traveled to for runner a to 100 meters, and for runner B to 90 meters. Solve for Tf in both equations and set them equal to each other. Using that, you get that Aa (acceleration of runner Aa = 10/9 * Ab That's great, and makes sense because runner A accelerates faster than runner B.

 

EDIT: Alternatively, if you're not allowed to use calc, you can calculate the area under the curve using the area of a triangle. It'll give you the same answer. Area of triangle = 1/2 * base * height. In our velocity vs. time graph, we get 1/2 * Time * velocity. We know from above that velocity for runner A = Aa*T, so using the triangle we get 1/2 * T * Aa * T or (Aa *T^2)/2, same as above. 

 

Now, we still have our distance equations from before. Da = T^2/2 * Aa and Db = T^2/2 * Ab. However, this time runner A has to run 110m, while runner B only has to run 100m.

 

So, using that info we could figure out at what time the two runners meet, but we have no good way of figuring out if that's before or after they finished the race. So, let's figure out the time it takes for each runner to run the 2nd race.

 

Da = 110 (runner A has to run 110 meters) = T^2/2 * Aa. Substitute in the Aa = 10/9 * Ab and you get 110=(5 * T^2 * Ab)/9. Solve for T^2 and you get T^2 = 198/Ab. That's important. Let's do the same for Db. Db = 100 = T^2/2 * Ab Solve for T^2 and you get T^2 = 200/Ab. Now, Ab is a constant, and T^2 is essentially T in this sense, so we can see that the time it takes for runner A to run 110 meters is very slightly less than the time it takes runner B to run 100 meters. 198/Ab < 200/Ab.

 

The interesting thing is that if instead of making runner A go back 10 meters, runner B went forward 10 meters, Runner B would have won. 

 

The only parabola in this question is in the position vs. time graph, which I didn't use. I guess you could have taken the derivative of the V vs. T graph to get position. So I guess there are few ways to solve the problem. 

 

Hope you're allowed to use Calc. 

[Trav_X]

11 hours ago, corrado33 said:

It's not too bad.

 

In the first race you know that Runner A runs 100m in some time (Tf). Runner B runs 90 meters in the same amount of time (Tf)

 

Using that, you can use the fact that the integral of the V vs. T graph equals the distance traveled. Therefore, if we can model the velocity like so

 

Va = Aa * T (Aa is the acceleration of runner A) and subsequently Vb = Ab * T for runner B.

 

Now, if we integrate (area under the curve) one of those to get the distance traveled we get Distance traveled = Aa ( T^2/2) So, we set the distance traveled to for runner a to 100 meters, and for runner B to 90 meters. Solve for Tf in both equations and set them equal to each other. Using that, you get that Aa (acceleration of runner Aa = 10/9 * Ab That's great, and makes sense because runner A accelerates faster than runner B.

 

EDIT: Alternatively, if you're not allowed to use calc, you can calculate the area under the curve using the area of a triangle. It'll give you the same answer. Area of triangle = 1/2 * base * height. In our velocity vs. time graph, we get 1/2 * Time * velocity. We know from above that velocity for runner A = Aa*T, so using the triangle we get 1/2 * T * Aa * T or (Aa *T^2)/2, same as above. 

 

Now, we still have our distance equations from before. Da = T^2/2 * Aa and Db = T^2/2 * Ab. However, this time runner A has to run 110m, while runner B only has to run 100m.

 

So, using that info we could figure out at what time the two runners meet, but we have no good way of figuring out if that's before or after they finished the race. So, let's figure out the time it takes for each runner to run the 2nd race.

 

Da = 110 (runner A has to run 110 meters) = T^2/2 * Aa. Substitute in the Aa = 10/9 * Ab and you get 110=(5 * T^2 * Ab)/9. Solve for T^2 and you get T^2 = 198/Ab. That's important. Let's do the same for Db. Db = 100 = T^2/2 * Ab Solve for T^2 and you get T^2 = 200/Ab. Now, Ab is a constant, and T^2 is essentially T in this sense, so we can see that the time it takes for runner A to run 110 meters is very slightly less than the time it takes runner B to run 100 meters. 198/Ab < 200/Ab.

 

The interesting thing is that if instead of making runner A go back 10 meters, runner B went forward 10 meters, Runner B would have won. 

 

The only parabola in this question is in the position vs. time graph, which I didn't use. I guess you could have taken the derivative of the V vs. T graph to get position. So I guess there are few ways to solve the problem. 

 

Hope you're allowed to use Calc. 

Thanks for the help  I understood most of your answers and how you got them, and I had gotten similar answers for the beginning part. The paragraph I need some more help with is this one

 

"Da = 110 (runner A has to run 110 meters) = T^2/2 * Aa. Substitute in the Aa = 10/9 * Ab and you get 110=(5 * T^2 * Ab)/9. Solve for T^2 and you get T^2 = 198/Ab. That's important. Let's do the same for Db. Db = 100 = T^2/2 * Ab Solve for T^2 and you get T^2 = 200/Ab. Now, Ab is a constant, and T^2 is essentially T in this sense, so we can see that the time it takes for runner A to run 110 meters is very slightly less than the time it takes runner B to run 100 meters. 198/Ab < 200/Ab"

 

The problem for me is just how you did the work on this problem. How did you get the (5 * T^2 * Ab)/9? I'm just not understanding your rationale for this part, probably cause I just don't know certain formulas that you're using  

[corrado33]

23 minutes ago, Trav_X said:

Thanks for the help  I understood most of your answers and how you got them, and I had gotten similar answers for the beginning part. The paragraph I need some more help with is this one

 

"Da = 110 (runner A has to run 110 meters) = T^2/2 * Aa. Substitute in the Aa = 10/9 * Ab and you get 110=(5 * T^2 * Ab)/9. Solve for T^2 and you get T^2 = 198/Ab. That's important. Let's do the same for Db. Db = 100 = T^2/2 * Ab Solve for T^2 and you get T^2 = 200/Ab. Now, Ab is a constant, and T^2 is essentially T in this sense, so we can see that the time it takes for runner A to run 110 meters is very slightly less than the time it takes runner B to run 100 meters. 198/Ab < 200/Ab"

 

The problem for me is just how you did the work on this problem. How did you get the (5 * T^2 * Ab)/9? I'm just not understanding your rationale for this part, probably cause I just don't know certain formulas that you're using  

Ok. The paragraph you're not understanding is how we figure out who won the second race.

 

Remember, from the first part we got that Aa = 10/9 * Ab. (Acceleration of runner A is 10/9s the acceleration of Runner B).

 

In the second race, runner A has to run 110 meters.

 

So, we substitute that into Runner A's distance equation. Da = T^2/2 * Aa. We got that equation either from the integral or by using the area of a triangle to solve for the area under the V vs. T graph for runner A. This is the "trick" for this problem. You have to know that the area under the velocity vs. time graph is equal to the distance traveled. The area under the graph is typically defined as the integral, but if you're not allowed to use calc the area under this velocity vs. time graph just happens to be a triangle. (Constant acceleration = a line with a positive slope on the V vs. T graph.) So you can also solve for the area (and therefore the distance covered) by simply using the area of a triangle. (1/2 * base * height.)

 

Now we stick 110 meters in: 110 = T^2/2 * Aa. 

 

Subsequently, for runner B: 100 = T^2/2 * Ab. 

 

Now, this doesn't help us. We still have 3 unknowns, but only two equations. So we substitute in the answer we got for the first part Aa = 10/9 * Ab into the first equation 110 = T^2/2 * Aa.

 

We get 110 = T^2/2 * (10/9 * Ab) or 110 = (10 * Ab * T^2)/(2 * 9)

 

That can simplify to 110 = (10 * Ab * T^2)/(18)

which can simplify to 110 = (5 * Ab * T^2)/(9) Which is what you see above. Looking back at it, I would have simplified it a bit differently, but that works. 

 

Now, solve the equation for T^2.

 

So now multiply by 9: 990 =  (5 * Ab * T^2)

divide by 5: 198 = Ab * T^2

Divide by Ab: 198 * Ab = T^2

 

 

[xentropa]

Distance = 1/2 * acceleration * time^2

 

Runner A runs 100 m in 10 sec.

That means 100 = 0.5 a(10)^2

or Runner A's acceleration (a) is 2 m/s^2

 

Runner B runs 90 m in 10 sec (since he is beaten by 10m)

That means 90 = 0.5 a (10)^2

or Runner B's acceleration is 1.8 m/s^2

 

so to figure out who wins if runner A gives a 10m head start, solve for ta and tb in

110 = 0.5 * 2 * (ta)^2

and

100 = 0.5 * 1.8 * (tb)^2

 

if ta (runner A time) > than tb (runner B time) then Runner A loses and vice versa.

[Trav_X]

32 minutes ago, corrado33 said:

Ok. The paragraph you're not understanding is how we figure out who won the second race.

 

Remember, from the first part we got that Aa = 10/9 * Ab. (Acceleration of runner A is 10/9s the acceleration of Runner B).

 

In the second race, runner A has to run 110 meters.

 

So, we substitute that into Runner A's distance equation. Da = T^2/2 * Aa. We got that equation either from the integral or by using the area of a triangle to solve for the area under the V vs. T graph for runner A. This is the "trick" for this problem. You have to know that the area under the velocity vs. time graph is equal to the distance traveled. The area under the graph is typically defined as the integral, but if you're not allowed to use calc the area under this velocity vs. time graph just happens to be a triangle. (Constant acceleration = a line with a positive slope on the V vs. T graph.) So you can also solve for the area (and therefore the distance covered) by simply using the area of a triangle. (1/2 * base * height.)

 

Now we stick 110 meters in: 110 = T^2/2 * Aa. 

 

Subsequently, for runner B: 100 = T^2/2 * Ab. 

 

Now, this doesn't help us. We still have 3 unknowns, but only two equations. So we substitute in the answer we got for the first part Aa = 10/9 * Ab into the first equation 110 = T^2/2 * Aa.

 

We get 110 = T^2/2 * (10/9 * Ab) or 110 = (10 * Ab * T^2)/(2 * 9)

 

That can simplify to 110 = (10 * Ab * T^2)/(18)

which can simplify to 110 = (5 * Ab * T^2)/(9) Which is what you see above. Looking back at it, I would have simplified it a bit differently, but that works. 

 

Now, solve the equation for T^2.

 

So now multiply by 9: 990 =  (5 * Ab * T^2)

divide by 5: 198 = Ab * T^2

Divide by Ab: 198 * Ab = T^2

 

 

Okay, that really helps  

 

I reread your paragraphs and you really helped me understand the problem better. Based on your work, is it impossible to solve exactly how much Runner A would beat Runner B by in the second race? Is that just completely dependent on how long it would take them to reach?

 

For an answer, I will use the Time^2 values (the 200/Ab and 198/Ab) and say once you plug in your T^2 values with seconds you can solve the problem for any possible amount of seconds.

[Trav_X]

I'm not sure how to get the distance between them when they finish the second race, that is all I am still left wondering about.