# Motion

• Oct 6th 2013, 10:06 PM
Fratricide
Motion
A car starts from rest and maintains a uniform acceleration of 2.0 ms-2 along a straight road. As the car starts, another car moving in the same direction along a prallel road with a constant velocity of 80 kmh-1 passes it. When will the cars be level with each other again?

I know that this is quite a simple question, but no matter what equations attempt to use I can't achieve the correct answer. I tried t = (v-u)/a, letting v = 22.22 ms-2, but got 11.11 s (which is incorrect). Does this mean that Car A was first passed by Car B after 11.11 seconds? Does this have any significance?

• Oct 6th 2013, 10:44 PM
mmonkman
Re: Motion
If the cars are parallel at t0, then the t you want is when their displacement is the same. What is d for a car with v0 and a=2.0ms^-2 ?
• Oct 6th 2013, 11:08 PM
Fratricide
Re: Motion
Quote:

Originally Posted by mmonkman
If the cars are parallel at t0, then the t you want is when their displacement is the same. What is d for a car with v0 and a=2.0ms^-2 ?

Do I need a value for t to solve this? If t = 0 and v = 0, then the equation for displacement simply equals 0 (i.e. they haven't begun moving yet).
• Oct 7th 2013, 12:07 AM
mmonkman
Re: Motion
No you do not want to input t=0. We already know that the cars are parallel at that time. What other value for t gets car A and car B with the same displacement. i.e. d of car A = d of car B. So you would want to solve for t.
• Oct 7th 2013, 12:21 AM
Fratricide
Re: Motion
What equation should I use?
• Oct 7th 2013, 12:37 AM
votan
Re: Motion
Quote:

Originally Posted by Fratricide
What equation should I use?

Because the initial velocity is 0:

d = (1/2)at^2 for the accelerated car
d = vt for uniform motion of the second car

Can you go from there?