Am I missing something? Basic acceleration

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Car A travels east at 15.7 m/s and crosses an intersection, while Car B starts from rest 42 m north of the intersection with a constant acceleration of 1.2 m/s². After 5 seconds, Car A is 78.5 meters east of the intersection, while Car B's displacement calculation initially confused the user due to incorrect use of initial velocity. The correct approach involves recognizing that Car B's initial velocity is zero, leading to a proper displacement calculation. The discussion also touches on finding the relative speed and acceleration of Car B with respect to Car A, emphasizing the need to set up equations based on their respective distances from the intersection.
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Hey everyone, quick question from the new guy. I don't know maybe I'm just making it harder than it is.

Car A is traveling east at 15.7 m/s. As car A crosses the intersection, car B starts from rest 42 m north of the intersection and moves south with a constant acceleration of 1.2 m/s2.
What is the distance between A and B 5 s after A crosses the intersection?

So car A is obviously 78.5 meters east of the intersection. It seems like a simple a^2 +b^2=c^2 problem but for some reason I can't convert car B's acceleration of 1.2 m/s^2 to a correct displacement. It seems to me car B should be 36m from the intersection but that can't be correct.
So can someone please explain to me how to convert acceleration to displacement.

P.S. I tried using delta x=vit+1/2axt2 with no luck either.
 
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jspek9 said:
I tried using delta x=vit+1/2axt2 with no luck either.
That's strange. I would have tried that. Could you show us the details of how you tried to use it?
 
I must be using it incorrectly. I used,
(1.2)*(5)+1/2(1.2)(5)^2= 21

then ((21)^2 +(78.5)^2)^1/2 = 81.3

Is my initial velocity wrong and what causing the problem?
 
jspek9 said:
I must be using it incorrectly. I used,
(1.2)*(5)+1/2(1.2)(5)^2= 21

then ((21)^2 +(78.5)^2)^1/2 = 81.3

Is my initial velocity wrong and what causing the problem?

What is the (1.2)*(5) ?
 
acceleration multiplied by time
 
jspek9 said:
acceleration multiplied by time

Yes. I know what it is. Just wondering why you put that term in there?

Wasn't car B at rest?
 
I see what your saying. The initial velocity isn't 1.2 its 0. That makes sense and gives a correct answer.

thanks for the help. This is the first time I have used this site and you guys where great.:!) Thanks again.
 
one more quick question though

the next question is asking what is the speed and magnitude of acceleration of car B relative to A

What exactly do they mean by relative?
 
In relation to each other, I'm guessing.
 
  • #10
just.karl said:
In relation to each other, I'm guessing.

Yes that is correct.
 
  • #11
can you be a little more specific? I can't seem to find anything in the text about how to go about seting these up.
 
  • #12
am I wrong assuming the final velociy of car B is 6 m/s at T=5?
 
  • #13
jspek9 said:
can you be a little more specific? I can't seem to find anything in the text about how to go about seting these up.

Start out by thinking about what is the equation for the distance car B travels relative to the origin (intersection) in terms of velocity and acceleration.

Likewise for the distance Car A travels from the origin (intersection).

Perhaps you can exploit the fact at any point in time the distance traveled by both is along the sides of a right triangle?
 
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