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Car motion

  • Thread starter Ry122
  • Start date
565
2
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?
 

LowlyPion

Homework Helper
3,079
4
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.
 
565
2
why is the squared variable multiplied by 1/2?
 

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