Setting two expressions equal to each other

[ImmmCanadian]

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

Homework Equations

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.

 

[DaveC426913]

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.)
Does that help you to reconcile the two equations together?

 

[nrqed]

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

Homework Equations

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 $a_B=-1 m/s^2$.

 

[Cutter Ketch]

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.

 

[lgg]

Your equations are correct, however, the (t-10) should be for car A, not B, as it starts ten seconds earlier. If you can't see it you can just set the time for car A to be t and for car B to be (t+10) as it will start moving after 10 seconds.

 

[kuruman]

This thread is more than 3 years old. Chances are that the original poster will not see your answer considering that he/she was last seen here 2 days after the original post. Thanks for your reply anyway and welcome to PF.

 

[nasu]

Your equations are correct, however, the (t-10) should be for car A, not B, as it starts ten seconds earlier. If you can't see it you can just set the time for car A to be t and for car B to be (t+10) as it will start moving after 10 seconds.

It is the other way around. For the car that starts 10 seconds later the time in the equations of motion should be t-10. At t=11s it only moved for 1 s and not for 21 seconds.