Simple Physics (Algebra) Question

[JustinSampson]

Hello,
I have (what should be) a simple physics question for you guys.

Say Point A and Point B are 500 miles apart. A car leaves Point A and a car leaves Point B at the same time. Car A is doing 50, Car B is doing 60.

How can I figure out where they will meet (in miles) from either Point A or B?

And this isn't my homework, in fact I'm old enough to be a lot of you guy's father's. It's just for an Flash game I'm working on, and can't remember how to do this.

Thanks,
Justin Sampson

 

[Lyfer]

Thats not physics.


Thats basic algebra.

 

[JustinSampson]

You're right now that I think of it. Changed!

 

[Lyfer]

Distance = Rate x Time


Now solve it.

 

[melvinfx]

294.11 they will collide

 

[JustinSampson]

Lyfer: I'm already that far, but thanks.

And melvinfx, care to share how you got that number? The numbers I gave aren't the actual ones I'm using (they'er variables).

 

[melvinfx]

Well i took the 50 mph and devided it by 500 = 10 hours. Then i took the 60 mph and devided it by 500 = 8.3 hours

so that is how long it would take to travel that distace.

Subtract the 10 and the 8.7 to get 1.7 that is the key!

Then I devided 500 with 1.7 to get 297.11

 

[CubicZirconia]

should read question closer next time

 

[hellfreeze]

when in doubt, just do it ghetto-status

50 --> 60
100 --> 120
150 --> 180

etc

 

[JustinSampson]

Great! Thanks for your help. Turns out I was doing it right (See, and you guys thought I was stupid! )

 

[oniq]

Originally posted by: hellfreeze
when in doubt, just do it ghetto-status

50 --> 60
100 --> 120
150 --> 180

etc

But hes programming so an equation or two would be more fitting to repeat the exercise.

 

[mee987]

(final position) = (speed)(time) + (initial position)
(in my coord. system car a is at x=0 and positive values to the right)
CAR B: x = 50*t + 0 = 50t
CAR A: x = -60*t + 500 = 500-60t

solve the system, t = ~4.5h
plug it into one of the equations, they collide 227 mi from where car a started.

at least i think thats right

 

[hellfreeze]

Originally posted by: oniq

Originally posted by: hellfreeze
when in doubt, just do it ghetto-status

50 --> 60
100 --> 120
150 --> 180

etc

But hes programming so an equation or two would be more fitting to repeat the exercise.

bah...like i said, when in doubt, just do it ghetto status

 

[JustinSampson]

Originally posted by: mee987
(final position) = (speed)(time) + (initial position)
(in my coord. system car a is at x=0 and positive values to the right)
CAR B: x = 50*t + 0 = 50t
CAR A: x = -60*t + 500 = 500-60t

solve the system, t = ~4.5h
plug it into one of the equations, they collide 227 mi from where car a started.

at least i think thats right

Perfect. Easy to code that. Thanks again Guys!

 

[kevinthenerd]

Originally posted by: mee987
(final position) = (speed)(time) + (initial position)
(in my coord. system car a is at x=0 and positive values to the right)
CAR B: x = 50*t + 0 = 50t
CAR A: x = -60*t + 500 = 500-60t

solve the system, t = ~4.5h
plug it into one of the equations, they collide 227 mi from where car a started.

at least i think thats right

you beat me to it. That's ALMOST the way I did it back in highschool... just spell everything out in terms of x1, x2, and t

to elaborate:

xB = xA

50t = 500 - 60t
110t = 500
t = 500 / 110 = 4.5454545454545...


but here's how I would have done it. I usually don't plug in the variables until the very end.

(assume 0, A, and B are subscripts. x, v, and a are main variables)

xB=xA

x0B + v0Bt + (1/2)aBt^2 = x0A + v0At + (1/2)aAt^2
(assuming constant acceleration)

x0B + v0Bt = x0A + v0At
(assuming no acceleration, as is your case)

v0Bt - v0At = x0A - x0B
t ( v0B - v0A ) = x0A - x0B
t = (x0A - x0B) / (v0B - v0A)
note that it's negative over negative. You can reverse it if desired:
t = (x0B - x0A) / (v0A - v0B)

I'll get back to you on finishing that the right way. Obligations call.

 

[her209]

d = r*t

for car starting from point A

d1 = r1*t

for car starting from point B

d2 = r2*t

where d1 + d2 = 500 when the two cars meet

so, d1 + d2 = r1*t + r2*t = (r1+r2)*t = 500

solving for t = 500 / 110

putting the value into one of the two equations above, you get:

d1 = 50*(500/110) = 227.272727.... miles from point A
-OR-
d2 = 60*(500/110) = 272.727272... miles from point B

where 227.272727 + 272.727272 = 500

 

[kevinthenerd]

That's good and efficient, but for the sake of higher physics I always like to have my final solution in terms of every variable. Instead of dropping in a previous answer (such as t in this case), I like putting the whole formula used to get t into the original equation. It gets to be pretty interesting as far as the algrebra goes in some problems.

 

[kevinthenerd]

plus, my formula allows greater flexibility. You can alter the starting positions, speeds, etc.

 

[RossGr]

OK lets write it out for you explicity.

Let the total distance = D
Let the Velocity of Car 1 = V
Let the velocity of Car 2 = W
Let the distance traveled by Car 1 = x
Let the distance traveled by Car 2 = y
Assuming that both cars start at the same time, they will have the same travel time = T


x = VT
y = WT

T= x/V = y/W

so x = yW/V

We have x+y=D
so
yW/V + y = D

y((W/V)+1) = D

y = D/(((W/V) +1)

or

y= DV/(W+V)

EDIT;
I have not done the algebra but I'll bet

x=DW/(W+V)

 

[dighn]

a really simple way of doing it: 500 miles apart, relative speed is 110. time to intercept = 500/110, then take the time and multiple by either speed and add to the corresponding starting position eg time*50 from A or time*60 from B