Relative Velocity and Acceleration

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Homework Help Overview

The problem involves two vehicles, Heather's Corvette and Jill's Jaguar, each with specified accelerations in a two-dimensional coordinate system. The task is to determine Heather's speed relative to Jill, the distance between them, and Heather's acceleration relative to Jill after a given time interval.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating relative velocity and acceleration using vector differences. Some express uncertainty about the correctness of their calculations and the application of kinematic equations.

Discussion Status

There is ongoing exploration of the relationships between acceleration, velocity, and distance. Some participants have suggested using kinematic equations, while others are questioning their calculations and seeking clarification on the concepts involved.

Contextual Notes

Participants are working under the constraints of starting from rest and using vector representations for acceleration and velocity. There is uncertainty regarding the application of kinematic principles to the problem setup.

pberardi
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Homework Statement



Heather in her Corvette accelerates at the rate of <3.20, -2.00>m/s^2, while Jill in her Jaguar accelerates at <1.00, 3.20>m/s^2. They both start from rest at the origin of an xy coordinate system. After 5.50 s, (a) what is Heather's speed with respect to Jill, (b) how far apart are they, and (c) what is Heather's acceleration relative to Jill?

Homework Equations





The Attempt at a Solution


Vhj = Vh - Vj = 3.77 - 3.35 = .421
[(3.77cos(327.9))i + (3.77sin(327.9))j] - [(3.35cos(72.6))i + 3.35sin(72.6))j
= 2.19i - 5.20j
magnitude is 5.64

I really don't know if this is correct and how it fits into the problem of getting the distance at 5.5 seconds and the speed at 5.50 seconds. Can someone help please?
 
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Once you determine the |acceleration| you can use ordinary kinematic means to determine d and v.

D = 1/2*|a|*t2

V = Vo + a*t

(but Vo is 0)
 
What about the speed? The answer .42 is incorrect. Did I calculate correctly?
 
pberardi said:
What about the speed? The answer .42 is incorrect. Did I calculate correctly?

I didn't check your math.

I wouldn't do it that way.

For V just multiply by the scalar 5.5 to get the Vx,y at 5.5 and then do the vector difference to determine the relative |v|.

Part c is the easiest to do because you just take the vector difference and you get the relative a(x,y). Then calculate |a| and apply the kinematic equation.
 
How do I get the acceleration? I would normally differentiate but I don't see how in this problem. Sorry for the bother.
 
pberardi said:
How do I get the acceleration? I would normally differentiate but I don't see how in this problem. Sorry for the bother.

Acceleration is a vector, just like velocity.

The relative acceleration then is the vector difference right?

Just like relative velocity is a vector difference. Speed is the scalar of V, or |V|.

The distance at any time can be found simply from kinematics by 1/2*|a|*t2 since they both started from rest.
 
I'm sorry I just don't get this. I do understand part c, but how to get part a and b in terms of kinematics and vectors is just too confusing for me to understand right now. I don't know why the answer .42 is incorrect. Can you try to explain it?
 
You are given the acceleration in x,y.

For Helen = 3.20 i - 2.0 j

Using V = a*t at 5.5 sec that makes Helen's velocity = 17.6 i - 11 j

Now figure Jill. And then figure your relative velocity.
 

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