When and where do the cars meet?

[CollegeStudent]

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?

 

[Doc Al]

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.)

 

[CollegeStudent]

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.

Well if the assumption I'm making based on your advice...I'd choose
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]

Well if the assumption I'm making based on your advice...I'd choose
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.

 

[CollegeStudent]

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 let's 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]

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 let's 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. )

 

[CollegeStudent]

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]

(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.