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Simple Physics (Algebra) Question

J

JustinSampson

Senior member
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: 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).
 
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
 
Originally posted by: hellfreeze
when in doubt, just do it ghetto-status

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

etc
Click to expand...

But hes programming so an equation or two would be more fitting to repeat the exercise.
 
(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
 
Originally posted by: oniq
Originally posted by: hellfreeze
when in doubt, just do it ghetto-status

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

etc
Click to expand...

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

bah...like i said, when in doubt, just do it ghetto status 😛
 
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
Click to expand...

Perfect. Easy to code that. Thanks again Guys!
 
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
Click to expand...


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.
 
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 🙂
 
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.
 
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)
 
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
 
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