Question

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?

Answer 1

How do we calculate for this

Answer 2

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.

Answer 3

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?

Answer 4

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?

Answer 5

Oh! That is for part c!

Answer 6

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.

Answer 7

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! :)

Answer 8

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

Answer 9

derinlerden what part is that for

Answer 10

thank you Eric

Answer 11

Thank you all for the help

Answer 12

a. 15 s
b. 30 m/s
c. 225 m

Answer 13

how did you get 30m/s

Answer 14

oh i get it thank you

Answer 15

I agree with all of those, derinlerden!

Answer 16

I'm glad you got it all PaaSolo!

Answer 17

onr more question im lost on how you calculated 15t=0.5at^2

Answer 18

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
\(t=\dfrac{15\ [\cancel m/\cancel s]}{ [\cancel m/s^{\cancel 2}]}=15\dfrac{1}{\dfrac{1}{[s]}}=15\ [s]\)

Answer 19

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 \(t\)!

Answer 20

thank you

Answer 21

You're very welcome!

Answer 22

Thanks also @theEric. I think I won't have to review after all, I will just print this and study it.

Answer 23

Haha, thank you! :)