Calculating Distance and Time for Meeting of Bicycle and Car

[bradyj7]

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

 

[Integral]

You will need to show us some of your thoughts and effort before we can help.

 

[bradyj7]

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

 

[SammyS]

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.

 

[bradyj7]

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

 

[vivekrai]

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