Velocity and acceleration

  • Thread starter bradyj7
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  • #1
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Hello,

Would somebody be able to explain to me how to calculate the answer to this question? I'm studying for an exam and cannot figure it out.

Question:

The rear of a bicycle passes a point O on a road with a velocity 4 ms-1 and an acceleration of 2 ms-2. Four seconds later the front of a car passes O with a velocity of 2 ms-1 and acceleration of 4 ms-2. When and how far from O does the front of the car meet the rear of the bicycle?

Many thanks

John
 

Answers and Replies

  • #2
Integral
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You will need to show us some of your thoughts and effort before we can help.
 
  • #3
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Hello,

This is what I think, but I can't seem to get the correct answer.

s1 = 4t + 0.5(2)t^2 distance bike goes
s2 = 2(t-4) + 0.5(4)(t-4)^2 distance car goes

then set s1 = s2 and solve

the answer is 16.5 seconds but I can't get this. maybe it is my algebra!

Thank you
 
  • #4
SammyS
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Hello,

This is what I think, but I can't seem to get the correct answer.

s1 = 4t + 0.5(2)t^2 distance bike goes
s2 = 2(t-4) + 0.5(4)(t-4)^2 distance car goes

then set s1 = s2 and solve

the answer is 16.5 seconds but I can't get this. maybe it is my algebra!

Thank you
That's the correct method.

Show how you solve for t.
 
  • #5
122
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I solved it and got the correct answer.

Originally I was using (t+4)

Could you explain to me why it is (t-4) and not (t+4)? What is reasoning behind it?

Thank you
 
  • #6
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It depends on for which motion do you take the time to be ' t '. See how does it make a difference?
 

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