Comparing Cars A and B: Who Wins the Race?

[4ertenok_1981]

Homework Statement

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance Da beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed Va. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed Vb, which is greater than Va.

How far from Car B's starting line will the cars be when Car B passes Car A?

Homework Equations

I have figured that the time to catch up = Da/(Vb-Va)

The Attempt at a Solution

Using the above, I now have this:

D=t(to catch up)*(Vb-Va)+Da

Does this look right?
thanks!

 

[learningphysics]

The time looks right. But not the distance.

 

[bel]

Consider car B, $$s = D_a + v_a t$$. Then consider car A, $$s = v_b t$$. However, the two $$s$$s are equal, and hence the equations could be equated. And you got that. Now going back to the distance, it is just $$s$$ at that specific time, and you already have $$s(t)$$ as a function of time $$t$$.

 

[4ertenok_1981]

Consider car B, $$s = D_a + v_a t$$. Then consider car A, $$s = v_b t$$. However, the two $$s$$s are equal, and hence the equations could be equated. And you got that. Now going back to the distance, it is just $$s$$ at that specific time, and you already have $$s(t)$$ as a function of time $$t$$.

...I don't understand the last part...

 

[bel]

The distance $$s$$ is the distance measured from the sarting point of car B. And that distance changes because the point where you measure it from are the cars and the cars are in motion. The car's motion is given as a function of a variable time $$t$$, which you found already.

 

[4ertenok_1981]

The distance $$s$$ is the distance measured from the sarting point of car B. And that distance changes because the point where you measure it from are the cars and the cars are in motion. The car's motion is given as a function of a variable time $$t$$, which you found already.

So, then...distance= t(to catch up)* (Vb-Va)?
:(
it's been a while since I took physics at high school, now I'm struggling at uni...

 

[learningphysics]

So, then...distance= t(to catch up)* (Vb-Va)?
:(
it's been a while since I took physics at high school, now I'm struggling at uni...

At any time t, what is the distance that car B has travelled?

 

[4ertenok_1981]

At any time t, what is the distance that car B has travelled?

v/t?

 

[bel]

Given the variable of time, $$t$$, you could use the either of the formulae describing the distance of the cars from the starting point of car B to find the distance, since at that time $$t$$, both the cars are at the same position. The main concern of this question is to reduce the number of variables as much as possible, so I've reduced it to only one distance variable $$s$$, to be measured from the initial position of car B and the variable $$t$$, which is the same for both cars and described both cars' motion with respect to these variables.

 

[learningphysics]

v/t?

Distance = velocity*time

I think we need to go back to the first part... look at bel's posts also... what is the position of car A at any time... what is the position of car B at any time...

 

[4ertenok_1981]

Given the variable of time, $$t$$, you could use the either of the formulae describing the distance of the cars from the starting point of car B to find the distance, since at that time $$t$$, both the cars are at the same position. The main concern of this question is to reduce the number of variables as much as possible, so I've reduced it to only one distance variable $$s$$, to be measured from the initial position of car B and the variable $$t$$, which is the same for both cars and described both cars' motion with respect to these variables.

I don't know why I'm not getting it...i'll try to start from scratch..
meanwhile, any sites with formulae you can recommend? I don't even have any of my high school notes.
Thanks so much for your help! :)

 

[4ertenok_1981]

Distance = velocity*time

I think we need to go back to the first part... look at bel's posts also... what is the position of car A at any time... what is the position of car B at any time...

Ok...Position of car A=v*t+Da
Position of car B-v*t...

 

[4ertenok_1981]

Ok...Position of car A=v*t+Da
Position of car B-v*t...

How about this?
distance= Vb*(Da/(Vb-Va)) ?
Or=Vb*t (time it took to catch up)?

 

[bel]

I already gave the formulae above, and by those (i.e., by equating the $$s$$s, as I mentioned), I came to your conclusion that "
I have figured that the time to catch up = Da/(Vb-Va)".

 

[learningphysics]

How about this?
distance= Vb*(Da/(Vb-Va)) ?

Yes, that's the answer. Be sure you understand why it's the answer though.

position of car A = Da + va*t
position of car B = vb*t

you set the two positions equal and solve for time. when you get the time, you can get the position, by either substituting into the car A equation or the car B equation. Both will work. Using the car B equation you immediately get vb*Da/(vb-va) I'll also do the car A one.

$$Da + va*(Da/(vb-va)) = \frac{Da*vb-Da*va +va*Da}{vb-va} = \frac{Da*vb}{vb-va}$$

the reason it works is because you solved for the time the positions are equal. so the two positions will come out the same with either formula.

It's better to use the car B formula since it's easier to evaluate the result.

 

[4ertenok_1981]

thanks!