# Velocity and acceleration

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

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
Staff Emeritus
Homework Helper
Gold Member
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.

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

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