# Setting equations equal to each other

## Homework Statement

Two cars are facing each other on a long straight airport runway. They are initially separated by a distance of 1 km. Car A begins to accelerate towards the other car at a uniform 0.5 ms^-2. Ten seconds later car B begins to move towards the other car with a uniform acceleration of 1 ms^-2. When and where do the two cars meet?

Car A:
u=0
v=?
a=0.5
t=t
s=1000-x

Car B:
u=0
v=?
a=1
t=t-10
s=x

s=vt
s=ut + 1/2at^2
s=vt-1/2at^2
v^2 = u^2 + 2as

## The Attempt at a Solution

1000-x= 0(t)+1/2(0.5)(t)^2
x= 0(t)+1/2(1)(t-10)^2[/B]
1000-x+x=.25t^2+0.5(t-10)^2
1000=.25t^2+0.5t^2-10t+50
0=.75t^2-10t-950
t=42.88 or -29.52
t=42.88
Which is wrong.

My confusion is coming from how to set equations equal to each other, not only in this problem.

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DaveC426913
Gold Member
I am not an expert in this, so I'm just talking here.

The one property both cars will have as equal will be t. It must be equal when they collide. (Provided you are including the 10 second lag, which it appears you are.)

nrqed
Homework Helper
Gold Member

## Homework Statement

Two cars are facing each other on a long straight airport runway. They are initially separated by a distance of 1 km. Car A begins to accelerate towards the other car at a uniform 0.5 ms^-2. Ten seconds later car B begins to move towards the other car with a uniform acceleration of 1 ms^-2. When and where do the two cars meet?

Car A:
u=0
v=?
a=0.5
t=t
s=1000-x

Car B:
u=0
v=?
a=1
t=t-10
s=x

s=vt
s=ut + 1/2at^2
s=vt-1/2at^2
v^2 = u^2 + 2as

## The Attempt at a Solution

1000-x= 0(t)+1/2(0.5)(t)^2
x= 0(t)+1/2(1)(t-10)^2[/B]
1000-x+x=.25t^2+0.5(t-10)^2
1000=.25t^2+0.5t^2-10t+50
0=.75t^2-10t-950
t=42.88 or -29.52
t=42.88
Which is wrong.

My confusion is coming from how to set equations equal to each other, not only in this problem.
It looks like a complicated way to do it, it would be easier to simply set the origin at the position of car A and just set x0=0 for car Am and set xo=1000 for car B and just set the final position xf of both cars equal.

But going back to your attempt, it looks like you forgot to set the acceleration of car B to be negative, it should be $a_B=-1 m/s^2$.

My confusion is coming from how to set equations equal to each other, not only in this problem.
You want the equations for each car to be written in one coordinate system. It doesn’t matter what you choose just so long as it is consistent. That includes using the sign consistently, for example right is always positive. (And that means one of the accelerations is negative)

Then you want to write both equations with a common clock, i.e. the same t. That means that one of the equations will be in terms of (t-10)

Use consistent coordinates and time base and you can then successfully find the place and time where the equations are equal.