Calculating Time for Car A to Pass Car B: Acceleration Comparison

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The discussion focuses on calculating the time it takes for Car A, which accelerates at 4.2 m/s², to pass Car B, which accelerates at 3.5 m/s² and starts 0.80 seconds earlier. The initial calculations provided by the user were incorrect due to the omission of the acceleration term and initial velocity for Car B. The correct approach involves using the equations of motion to set the distances equal and solving for time, leading to a quadratic equation that must be solved to find the intersection point.

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Car A accelerates at 4.2ms
Car B accelerates at 3.5ms
In a drag Car B leaves the start line .80 secs earlier than car A.
At what time does car A pass car B?
My attempt:
Car A:
y=4.2mx

Determine constant for car B
v=u+at
v=0+(3.5)(.80)=2.8
s=1/2(0+2.8).80
s=1.12metres

Car B:
y=3.5x + 1.12

3.5x + 1.12 = 4.2x
to find when graph intercepts
-0.7x=-1.12
x=1.6 seconds
What am I doing wrong?
 
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Ry122 said:
Car B:
y=3.5x + 1.12

What am I doing wrong?

For one thing you have apparently forgotten the acceleration term.
For the other you have forgotten what your initial velocity was.

Xb = 1.12 + 2.8t + 1/2*3.5*t2

Xa = 1/2*4.2*t2

When the distances are the same then t is your time.

Of course it will require a quadratic solution.
 
why is the squared variable multiplied by 1/2?
 

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