anonymous
  • anonymous
Car A is stopped at a red light . At the instant the light turns green, Car B comes speeding past in the adjacent lane with a constant speed pf 15m/s. Car A begins to accelerate at a rate of 2m/s^2. zhow long does it take for Car A to catch up with Car B? b. How fast is Car Amoving when it catxhes up with Car B? c. How far have the two cars travelled from the traffic light by the time Car A catches up with Car B?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
How do we calculate for this
radar
  • radar
I would need to review quite a bit here, but I sure would like to see the solution even if, I don't think I could do it now.
theEric
  • theEric
Hi! What is true when the two cars pass each other? The distance they traveled from the light is the same! So, if you find the distances they traveled, you set them equal. Do you know how functions work?

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More answers

theEric
  • theEric
Car A has a constant acceleration, we'll say it's \(a_A\) for the speaking math. Car B has a constant velocity, we'll say it's \(v_B\). And so, what is the distance function of each? Like you have an expression that's equal to distance, and has some variables maybe? I don't know if you're familiar, but there is the formula \(x=x_0+v_0\ \Delta t+\frac{1}{2}a\ (\Delta t)^2\). Do you know everything in that formula?
theEric
  • theEric
Oh! That is for part c!
theEric
  • theEric
You solve for \(t\) for part A. For part B, since you have the time interval from part A, you can use the definition of acceleration: \(a=\dfrac{v-v_0}{\Delta t}\) and solve for \(v\) since you know the other values.
theEric
  • theEric
Good luck! Ask questions if you need! If the framework I laid out is correct, many people will be able to help in this section! :)
anonymous
  • anonymous
15t for the constant speed 0.5a*t^2 for the accelerating speed 15t=0.5a*t^2 then t=15 sec 15*15=225 m
anonymous
  • anonymous
derinlerden what part is that for
anonymous
  • anonymous
thank you Eric
anonymous
  • anonymous
Thank you all for the help
anonymous
  • anonymous
a. 15 s b. 30 m/s c. 225 m
anonymous
  • anonymous
how did you get 30m/s
anonymous
  • anonymous
oh i get it thank you
theEric
  • theEric
I agree with all of those, derinlerden!
theEric
  • theEric
I'm glad you got it all PaaSolo!
anonymous
  • anonymous
onr more question im lost on how you calculated 15t=0.5at^2
theEric
  • theEric
Okay! Well, I was vague earlier. But I said that "What is true when the two cars pass each other? The distance they traveled from the light is the same!" And I went into saying about finding the distance functions. Well, look at \(x=x_0+v_0\ t+\frac{1}{2}a\ t^2\). For both, \(x_0=0\) since they start at the beginning and it's convenient to say that the stoplight is at \(x=0\). So \(x=v_0\ t+\frac{1}{2}a\ t^2\). For car A, which is accelerating, it's \(v_0=0\), since it starts from a stop. It's \(a_A=2\ [m/s^2]\). So \(x_A=\frac{1}{2} 2\ [m/s^2]\ t^2=[m/s^2]\ t\). For car B, which has \(a=0\) but \(v_0=15\ [m/s]\). So \(x_B=15\ [m/s]\ t\). Those two distances, of A and B (\(x_A\) and \(x_b\)), are equal when the cars pass! So \(x_A=x_B\). By a substitution, \(2\ [m/s^2]\ t^2=15\ [m/s]\ t\). Then.... solve for \(t\). \( [m/s^2]\ t^2=15\ [m/s]\ t\) \(\qquad\Downarrow\qquad\)divide both sides by \(t\) \([m/s^2]\ t=15\ [m/s]\) \(\qquad\Downarrow\qquad\)divide both sides by \([m/s^2]\) \(t=\dfrac{15\ [m/s]}{ [m/s^2]}\) \(\qquad\Downarrow\qquad\)simplify \(t=\dfrac{15\ [\cancel m/\cancel s]}{ [\cancel m/s^{\cancel 2}]}=15\dfrac{1}{\dfrac{1}{[s]}}=15\ [s]\)
theEric
  • theEric
Oh! I see I wrote \(2\ [m/s^2]\ t^2=15\ [m/s]\ t\) at one point. That should just be \([m/s^2]\ t^2=15\ [m/s]\ t\) like how I started solving for \(t\)!
anonymous
  • anonymous
thank you
theEric
  • theEric
You're very welcome!
radar
  • radar
Thanks also @theEric. I think I won't have to review after all, I will just print this and study it.
theEric
  • theEric
Haha, thank you! :)

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