# When and where do the cars meet?

## Homework Statement

Two traffic lights are 100m apart. Light 1 being due west from light 2. Car 1 is moving east at a constant speed of 25m/s. When car 1 passes light 1, car 2 starts from rest, west at a constant acceleration of 2.0 m/s². Where do they pass and how long after the light changes do they pass?

## Homework Equations

I was thinking Vf² = Vi² + 2aΔx

## The Attempt at a Solution

Well I tried solving for the final velocity of car 2, since Vf of car 1 would be the same as initial.

Vf² = Vi² + 2aΔx
= 0² + 2(2.0m/s/s)(100)
Vf² = 400
Vf = 20

I'm not sure if this was the right approach but can someone point me in the right direction here?

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Doc Al
Mentor
I assume car 2 starts out from light 2?

See if you can express the position of each car as a function of time. (Hint: Measure the position of each from the same point.)

I assume car 2 starts out from light 2?

See if you can express the position of each car as a function of time. (Hint: Measure the position of each from the same point.)
Yeah sorry it starts from light 2.

Δx = 1/2(Vf + Vi)t for the formula.

So for time I believe it would be

t = (2Δx)/(Vf+Vi)

I decided to measure from 50m

so

Car 1 = 2(50)/50 = 2 seconds to reach 50m

and
assuming I calculated Vf correctly up in the original question (but subbing in 50 for 100)

Car 2 = 2(50)/14.14213562 = 7.071067814 seconds to reach 50m

So now do I just set these 2 equal to each other?

25 m/s = 7.071067814 m/s

divide both sides by 7.071067814 making the time it takes to pass 3.5 s?

Doc Al
Mentor
Δx = 1/2(Vf + Vi)t for the formula.
That's OK, but realize that Vf also depends on time.

You'd be better off using:
$$x = x_0 + v_0 t + (1/2) a t^2$$

That's what I had in mind when I spoke of position as a function of time.

Note: The calculation you made for Vf in your first post is not quite relevant. You found the speed of car 2 when it reaches light 1, not when it passes car 1.

You'd be better off using:
$$x = x_0 + v_0 t + (1/2) a t^2$$

That's what I had in mind when I spoke of position as a function of time.
Ahh, see I'm going off an equation sheet my prof gave us
Actually:
Δx = Vi*t + (1/2) a t^2...could this also be written as:
X - Xo = Vi*t + (1/2) a t^2
X = Vi*t + (1/2) a t^2 + X_o

Yeah, that makes sense...okay so lets see

Car 1 would be 25m/s * t
Car 2 would be (1/2)(-2.0m/s/s) t^2 + 100m
so
(-1m/s/s) t^2 + 100m

So NOW set them equal to each other?

(-1m/s/s) t^2 + 100m = 25m/s/s t

(-1m/s/s) t^2 - 25m/s/s t + 100m = 0

Ahh a quadratic...hmm well after the calculating I get -28.5 , 3.51

Obviously time can't be negative so I'm assuming 3.51 would be the time they intersect??

Doc Al
Mentor
Ahh, see I'm going off an equation sheet my prof gave us
Actually:
Δx = Vi*t + (1/2) a t^2...could this also be written as:
X - Xo = Vi*t + (1/2) a t^2
X = Vi*t + (1/2) a t^2 + X_o

Yeah, that makes sense...okay so lets see

Car 1 would be 25m/s * t
Car 2 would be (1/2)(-2.0m/s/s) t^2 + 100m
so
(-1m/s/s) t^2 + 100m

So NOW set them equal to each other?

(-1m/s/s) t^2 + 100m = 25m/s/s t

(-1m/s/s) t^2 - 25m/s/s t + 100m = 0

Ahh a quadratic...hmm well after the calculating I get -28.5 , 3.51

Obviously time can't be negative so I'm assuming 3.51 would be the time they intersect??
Looks good. (I'll have to have a second look at what you did before, since you somehow got that same answer. )

Looks good. (I'll have to have a second look at what you did before, since you somehow got that same answer. )
Yeah I noticed that as well...however in terms of significant figures...

25 / 7.071067814 comes out to 3.535533905 with 3 sig figs = 3.54

and this other method came to 3.507810594 and here would be 3.51

So with the 3 sig figs it would make a difference...not much at all...but different

Doc Al
Mentor
(I'll have to have a second look at what you did before, since you somehow got that same answer. )
I did take a second look at your earlier work and I cannot understand your reasoning. Probably just a fluke that you got the "right" answer that way.